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Team PrabhaT

15 Mock testcoMBINeD

pHYsIcs, cHeMIstrY AND MAtHeMAtIcs

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IIT-JEE JEE MAIN AND ADVANCED15 MoCk TEsT

CoMBINEDpHYsICs, CHEMIsTrY AND MATHEMATICs

by Team Prabhat

Published by prABHAT prAkAsHAN4/19 Asaf Ali Road, New Delhi–110 002

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l Mock Test – 1 Physics ...................................................................................................................................1-18

l Mock Test – 2 Physics .................................................................................................................................19-32




l Mock Test – 1 Chemistry ............................................................................................................................73-91

l Mock Test – 2 Chemistry ..........................................................................................................................92-102

l Mock Test – 3 Chemistry ........................................................................................................................103-113



l Mock Test – 1 Mathematics ....................................................................................................................135-154





CONTENTS

1Mock Test-1

Mock Test “JEE-Main”

Do not open this Test Booklet until you are asked to do so.

Read carefully the Instructions on the Back Cover of this Test Booklet.

Important Instructions:

1. Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball Point Pen. Use of pencil is strictly

prohibited.

2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer

Sheet and fill in the particulars carefully.

3. The test is of 3 hours duration.

4. The Test Booklet consists of 75 questions. The maximum marks are 300.

5. There are three parts in the question paper A, B, C consisting of, Physics, Chemistry and Mathematics having 25

questions in each part of equal weightage.

6. Each question is allotted 4 (four) marks for each correct response. For MCQs - 4 Marks will be awarded for every correct

answer and 1 Mark will be deducted for every incorrect answer.

7. For answer with numeric value - 4 Marks will be awarded for every correct answer and 0 Mark will be deducted for every

incorrect answer.

8. Use Blue/Black Ball Point Pen only for writing particulars/ marking responses on Side-1 and Side-2 of the Answer Sheet.

Use of pencil is strictly prohibited.

9. No candidates is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any

electronic device, etc., except the Admit Card inside the examination hall/room.

10. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the

bottom of each page and at the end of the booklet.

11. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room / Hall.

However, the candidates are allowed to take away this Test Booklet with them.

12. The CODE for this Booklet is A. Make sure that the CODE printed on Side-2 of the Answer Sheet and also tally the serial

number of the Test Booklet and Answer Sheet are the same as that on this booklet. In case of discrepancy, the candidate

should immediately report the matter to the invigilator for replacement of both the Test Booklet and the Answer Sheet.

13. Do not fold or make any stray marks on the Answer Sheet.

Name of the Candidate (in Capital letters):

Roll Number : in figures

in words

Examination Centre Number:

Name of Examination Centre (in Capital letters):

Candidate’s Signature: Invigilator’s Signature:

A Test Booklet Code

Physics 2

Read the following instructions carefully:

1. The candidates should fill in the required particulars on the Test Booklet and Answer Sheet (Side-1) with

Blue/Black Ball Point Pen.

2. For writing/marking particulars on Side-2 of the Answer Sheet, use Blue/Black Ball Point Pen only.

3. The candidates should not write their Roll Numbers anywhere else (except in the specified space) on the Test

Booklet/Answer Sheet.

4. Out of the four options given for each question, only one option is the correct answer.

5. For each incorrect response, one-fourth (¼) of the total marks allotted to the question would be deducted for

MCQs and No deduction for numeric questions from the total score. No deduction from the total score,

however, will be made if no response is indicated for an item in the Answer Sheet.

6. Handle the Test Booklet and Answer Sheet with care, as under no circumstances (except for discrepancy in

Test Booklet Code and Answer Sheet Code), another set will be provided.

7. The candidates are not allowed to do any rough work or writing work on the Answer Sheet. All

calculations/writing work are to be done in the space provided for this purpose in the Test Booklet itself,

marked ‘Space for Rough Work’. This space is given at the bottom of each page and at the end of the booklet.

8. On completion of the test, the candidates must hand over the Answer Sheet to the Invigilator on duty in the

Room/Hall. However, the candidates are allowed to take away this Test Booklet with them.

9. Each candidate must show on demand his/her Admit Card to the Invigilator.

10. No candidate, without special permission of the Superintendent or Invigilator, should leave his/her seat.

11. The candidates should not leave the Examination Hall without handing over their Answer Sheet to the

Invigilator on duty and sign the Attendance Sheet again. Cases where a candidate has not signed the Attendance

Sheet a second time will be deemed not to have handed over the Answer Sheet and dealt with as an unfair

means case. The candidates are also required to put their left hand THUMB impression in the space

provided in the Attendance Sheet.

12. Use of Electronic/Manual Calculator and any Electronic Item like mobile phone, pager etc. is prohibited

13. The candidates are governed by all Rules and Regulations of the JAB/Board with regard to their conduct in the

Examination Hall. All cases of unfair means will be dealt with as per Rules and Regulations of the JAB/Board.

14. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.

15. Candidates are not allowed to carry any textual material, printed or written, bits of papers, pager,

mobile phone, electronic device or any other material except the Admit Card inside the examination

hall/room.

3 Mock Test-1

JEE-MAIN: PHYSICS MOCK TEST-1

SECTION 1 (Multiple Choice Question)

1. If a particle moves from point P (2,3,5) to point Q (3,4,5).

Its displacement vector be:

a. ˆˆ ˆ 10i j k+ + b. ˆˆ ˆ 5i j k+ +

c. ji ˆˆ + d. ˆˆ ˆ2 4 6i j k+ +

2. Long horizontal rod has a bead which can slide along its

length and is initially placed at a distance L from one end

A of the rod. The rod is set in angular motion about A with

a constant angular acceleration α. If the coefficient of

friction between the rod and bead is µ, and gravity is

neglected, then the time after which the bead starts

slipping is:

a.

µ

α b.

µ

α

c.

1

µα d. infinitesimal

3. A wind-powered generator converts wind energy into

electric energy. Assume that the generator converts a

fixed fraction of the wind energy intercepted by its blades

into electrical energy. For wind speed v, the electrical

power output will be proportional to

a. v b. v2

c. v3 d. v

4

4. A spherical hollow is made in a lead sphere of radius R,

such that its surface touches the outside surface of lead

sphere and passes through the center. What is the shift in

the center of mass of lead sphere due to the hollowing?

a. 7

R b.

14

R

c. 2

R d. R

5. The bob of a simple pendulum executes simple harmonic

motion in water with a period t, while the period of

oscillation of the bob is 0t in air. Neglect frictional force

of water and given that the density of the bob is

(4 /3) 1000× kg/m3. What relationship is true between t

and 0t ?

a. 0t t= b. 0 / 2t t=

c. 02t t= d. 04t t=

6. A capillary tube A is dipped in water. Another identical

tube B is dipped in a soap-water solution. Which of the

following shows the relative nature of the liquid columns

in the two tubes?

a. b.

c. d.

7. Two rods of different materials having coefficients of

linear expansion 1 2,α α and Young’s

moduli 1 2,Y Y respectively are fixed between two rigid

massive walls. The rods are heated such that they undergo

the same increase in temperature. There is no bending of

the rods. If 1 2: 2 : 3,α α = the thermal stresses developed

in the two rods are equal provided 1 2:Y Y is equal to:

a. 2 : 3 b. 1 : 1

c. 3 : 2 d. 4 : 9

8. Three liquids of equal volumes are thoroughly mixed. If

their specific heats are 1 2 3, ,s s s and their temperature

1 2 3, ,θ θ θ and their densities 1 2 3, ,d d d respectively then

the final temperature of the mixture is:

a. 1 1 2 2 3 3

1 1 2 2 3 3

s s s

d s d s d s

θ θ θ+ ++ +

b. 1 1 1 2 2 2 3 3 3

1 1 2 2 3 3

d s d s d s

d s d s d s

θ θ θ+ ++ +

c. 1 1 1 2 2 2 3 3 3

1 1 2 2 3 3

d s d s d s

d d d

θ θ θθ θ θ

+ ++ +

d. 1 1 2 2 3 3

1 1 2 2 3 3

d d d

s s s

θ θ θθ θ θ

+ ++ +

9. Two monatomic ideal gases 1 and 2 of molecular masses

1m and

2m respectively are enclosed in separate containers

kept at the same temperature. The ratio of the speed of

sound in gas 1 to the gas 2 is given by:

a. 1

2

m

m b. 2

1

m

m

c. 1

2

m

m d. 2

1

m

m

10. A gas is compressed adiabatically till its temperature is

doubled. The ratio of its find volume to initial volume will

be:

a. 1

2 b. more than

1

2

c. less than 1

2 d. between 1and 2

A B A B

A B B A

Physics 4

11. A black body is at a temperature of 2880 K. The energy of

radiation emitted by this body with wavelength between

499 nm and 500 nm is 1,U between 999 nm and 1000 nm

is 2U and between 1499 nm and 1500 nm is

3.U The

Wien constant, 62.88 10 nm-K.= ×b Then,

a. 1 0=U b.

3 0=U

c. 1 2>U U d.

2 1>U U

12. In moving from A to B along an electric field line, the

electric field does 196.4 10−× J of work on an electron. If

1 2,φ φ are equipotential surfaces, then the potential difference

( )C AV V− is

a. – 4V b. 4V

c. Zero d. 64 V

13. Force acting on a charged particle kept between the plates

of a charged condenser is F. If one of the plates of the

condenser is removed, the force acting on the same

particle becomes:

a. 0 b. 2

F

c. F d. 2F

14. An electric cable of copper has just one wire of radius

9 mm. Its resistance is 50 Ω. This single copper wire of

cable is replaced by 6 different well insulated copper

wires, each of radius 3 mm. The total resistance of the

cable will now be equal to:

a. 7.5 Ω b. 45 Ω

c. 90 Ω d. 270 Ω

15. An infinitely long conductor PQR is bent to form a right

angle as shown. A current I flows through PQR. The

magnetic field due to this current at the point M is H1.

Now another infinitely long straight conductor QS is

connected at Q so that the current is I/2 in QR as well as in

QS, The current in PQ remaining unchanged. The

magnetic field at M is now .2H .The ratio 21 /HH is given

by

a. 1

2 b. 1

c. 2

3 d. 2

16. Double slit interference experiment is carried out with

monochromatic light and interference fringes are observed. If

now monochromatic light is replaced by white light, what

change is expected in interference pattern?

a. no change

b. pattern disappears

c. white and dark fringes are observed throughout the

pattern

d. a few coloured fringes are observed on either side of

central white fringe

17. The electric field of an electromagnetic wave in free space

is given by 7 ˆ10cos (10 ) V/m,E t kx j= + Where, t and x are

in seconds and metres respectively. It can be inferred that

(a) the wavelength λ is 188.4 m

(b) the wave number k is 0.33 rad/m

(c) the wave amplitude is 10 V/m

(d) the wave is propagating along + x direction

Which one of the following pairs of statements is correct?

a. (c) and (d) b. (a) and (b)

c. (b) and (c) d. (a) and (c)

18. Light of wavelength λ strikes a photosensitive surface

and electrons are ejected with kinetic energy E. If the

kinetic energy is to be increased to 2E, the wavelength

must be changed to λ′ where:

a. 2

λλ ′= b. 2λ λ′=

c. 2

λλ λ′< < d. λ λ′=

19. The ionisation potential of hydrogen atom is 13.6 V. The

energy required to remove an electron in the n = 2 state of

the hydrogen atom is

a. 27.2 eV b. 13.6 eV

c. 6.8 eV d. 3.4 eV

20. Hydrogen bomb is based on which of the following

phenomenon?

a. Nuclear fission

b. Nuclear fusion

c. Radioactive decay

d. None of these

90o

90o

M

Q P – ∞

S + ∞

– ∞

R

I

5 Mock Test-1

SECTION 2 (Numeric Value Question)

21. A cube has a side of length 21.2 10−× m. Calculate its

volume:

a. 6 31.7 10 m−× b. 6 31.73 10 m−×

c. 6 31.70 10 m−× d. 6 31.732 10 m−×

22. A body of mass 2 kg is released at the top of a smooth

inclined plane having inclination 30 .° It takes 3 seconds to

reach the bottom. If the angle of inclination is doubled

keeping same height (i.e., made 60° ), what will be the

time taken?

a. 3 sec b. 3 sec

c. 3 3 sec d. 1.5 sec

23. There are two bodies of masses 100 kg and 10000 kg

separated by a distance 1m. At what distance from the

smaller body, the intensity of gravitational field will be

zero:

a. 1

9m b.

1

10m

c. 1

11 m d.

10

11m

24. Due to a small magnet intensity at a distance x in the end

on position is 9 gauss. What will be the intensity at a

distance 2

x on broad side on position?

a. 9 gauss b. 4 gauss

c. 36 gauss d. 4.5 gauss

25. When both source and observe approach each other with a

velocity equal to the half the velocity of sound the change

in frequency of sound as detected by the listener is:

a. 0 b. 25 %

c. 50% d. 150%

Space for rough work

Physics 6

JEE ADVANCE PAPER-I

Time 3 Hours. Max. Marks 180 (60 for Physics)

Read the Instructions Carefully

Question Paper Format and Marking Scheme:

1. The question paper has three parts: Physics, Chemistry and Mathematics. Each part has three sections.

2. Section 1 contains 6 multiple choice questions with one or more than one correct option.

Marking Scheme: +4 if only (all) the correct option(s) is (are) chosen, +3 if all the four options are correct but only three

options are chosen, +2 if three or more options are correct but only two options are chosen, both of the options must be

correct, +1 if two or more options are correct buy only one option is chosen and it must be correct, -2 (in all other cases).

3. Section 2 contains 8 Numerical value answer type questions. The answer to each question is a single digit integer ranging

from 0 to 9 (both inclusive).

Marking Scheme: +3 if only correct numerical value is given.

4. Section 3 contains 2 Paragraph based questions (2 paragraphs, each having 2 MCQs with one correct answer only)

Marking Scheme: +3 if only the correct option is selected, -1 in all other cases.

Note:

Possible response Section 1 Section 2 Section 3

Not attempted 0 0 0

Partial Correct +1 for each correct option selected 0 0

Correct +4 +3 3

incorrect -2 0 -1

SECTION 1 (Maximum Marks: 24)

MCQs with one or more than one correct answer

1. A transparent thin film of uniform thickness and refractive

index 1 1.4=n is coated on the convex spherical surface of

radius R at one end of a long solid glass cylinder of

refractive index 2 1.5,=n as shown in the figure. Rays of

light parallel to the axis of the cylinder traversing through

the film from air to glass get focused at distance 1f from

the film, while rays of light traversing from glass to air get

focused at distance 2f from the film. Then

a. 1| | 3=f R b.

1| | 2.8=f R

c. 2| | 2=f R d.

2| | 1.4=f R

2. A conductor (shown in the figure) carrying constant

current I is kept in the x-y plane in a uniform magnetic

field .B

If F is the magnitude of the total magnetic force

acting on the conductor, then the correct statement(s)

is(are)

a. If B

is along ˆ, ( )z F L R∝ +

b. If B

is along ˆ, 0x F =

c. If B

is along ˆ, ( )y F L R∝ +

d. If B

is along ˆ, 0z F =

3. Planck’s constant h, speed of light c and gravitational

constant G are used to form a unit of length L and a unit of

mass M. Then the correct option(s) is(are)

a. M c∝ b. M G∝

c. L h∝ d. L G∝

4. Two independent harmonic oscillators of equal mass are

oscillating about the origin with angular frequencies 1ω

and 2ω and have total energies

1E and 2,E respectively.

The variations of their momenta p with positions x are

/6π /4πL R R L

R R y

x

n1

n2 Air

7 Mock Test-1

shown in the figures. If 2an

b= and ,

an

R= then the correct

equation(s) is(are)

a. 1 1 2 2E Eω ω= b. 22

1

nωω

=

c. 2

1 2nω ω = d. 1 2

1 2

E E

ω ω=

5. Inner and outer radii of a spool are r and R respectively.

A thread is wound over its inner surface and spool is

placed over a rough horizontal surface. Thread is pulled

by a force F as shown in figure. In case of pure rolling,

which of the following statements are false?

a. Thread unwinds, spool rotates anticlockwise and

friction acts leftwards

b. Thread winds, spool rotates clockwise and friction acts

leftwards

c. Thread winds, spool moves to the right and friction acts

rightwards

d. Thread winds, spool moves to the right and friction

does not come into existence

6. Two ideal batteries of emf 1V and

2V and three resistances

1 2,R R and 3R are connected as shown in the figure. The

current in resistance R2 would be zero if

a. 1 2=V V and 1 2 3= =R R R

b. 1 2=V V and

1 2 32= =R R R

c. 1 22=V V and

1 2 32 2= =R R R

d. 1 22 =V V and

1 2 32 = =R R R


Numerical value answer type questions

7. A small filament is at the centre of a hollow glass sphere

of inner and outer radii 8 cm and 9 cm respectively. The

refractive index of glass is 1.50. Calculate the position of

the image of the filament when viewed from outside the

sphere.

8. An infinitely long uniform line charge distribution of

charge per unit length λ lies parallel to the y-axis in the

y-z plane at3

z2a= (see figure). If the magnitude of the

flux of the electric field through the rectangular surface

ABCD lying in the x-y plane with its center at the origin is

0

L

n

λε

0(ε = permittivity of free space), then the value of n

is

9. For an atom of an ion having single electron, the

wavelength observed1 2λ = are units and

3 3λ = units

figure. The value of missing wavelength 2λ is

10. Consider an elliptically shaped rail PQ in the vertical

plane with 3 mOP = and 4 m.OQ = A block of mass

1 kg is pulled along the rail from P to Q with a force of

18 N, which is always parallel to line PQ (see the given

figure). Assuming no frictional losses, the kinetic energy

1λ

n3 orbit

n2 orbit

n1 orbit

3λ

2λ

C

z

D L

O

B A

x

a

3

2a

y

Glass

O B Air

OA = 8 cm

OB = 9 cm

m = 1.50

2V

1V

1R

2R

3R

r

R α

F

Energy = E1 P

a

b x x

R

P Energy = E2

A

Physics 8

of the block when it’s reaches Q in ( 10)×n joules. The

value of n is (take acceleration due to gravity210 ms−)

11. Two identical uniform discs roll without slipping on two

different surfaces AB and CD (see figure) starting at A and

C with linear speeds 1v and

2 ,v respectively, and always

remain in contact with the surfaces. If they reach B and D

with the same linear speed and 1 3 / ,v m s= then

2v in m/s

is 2( 10 m / s )g =

12. Two spherical stars A and B emit blackbody radiation. The

radius of A is 400 times that of B and A emits 410 times

the power emitted from B. The ratio ( / )A Bλ λ of their

wavelengths Aλ and

Bλ at which the peaks occur in their

respective radiation curves is

13. The half life of a freshly prepared radioactive sample is 2 h.

If the sample emits radiation of intensity, which is 16 times

the permissible safe level, then the minimum time taken

after which it would be possible to work safely with source

is

14. A Young’s double slit interference arrangement with slits

1S and 2S is immersed in water (refractive index = 4/3) as

shown in the figure. The positions of maxima on the

surface of water are given by 2 2 2 2 2 ,x p m dλ= − where λ

is the wavelength of light in air (refractive index = 1), 2d

is the separation between the slits and m is an integer. The

value of p is


2 Paragraph based questions (2 paragraphs, each having

2 MCQs with one correct answer only)

Paragraph for Question No. 15 to 16

Light guidance in an optical fiber can be understood by

considering a structure comprising of thin solid glass cylinder

of refractive index1n surrounded by a medium of lower

refractive index 2 .n

The light guidance in the structure takes

place due to successive total internal reflections at the interface

of the media 1n and

2n as shown in the figure. All rays with

the angle of incidence i less than a particular value mi are

confined in the medium of refractive index 1.n The numerical

aperture (NA) of the structure is defined as sin .mi

15. For two structures namely 1S with

1 45 / 4n = and

2 3/ 2,n = and 2S with

1 8 / 5n = and 2 7 / 5n = and taking

the refractive index of water to be 4/3 and that of air to be

1, the correct option(s) is(are)

a. NA of 1S immersed in water is the same as that of 2S

immersed in a liquid of refractive index 16

3 15

b. NA of 1S immersed in liquid of refractive index

6

15

is the same as that of 2S immersed in water

c. NA of 1S placed in air is the same as that of

2S

immersed in liquid of refractive index 4

.15

d. NA of 1S placed in air is the same as that of

2S placed

in water

16. If two structures of same cross-sectional area, but different

numerical apertures1NA and

2 2 1( )NA NA NA< are joined

longitudinally, the numerical aperture of the combined

structure is

1nθ

2n

1 2n n>

Air Cladding

i Core

xd

2S

d

1S

Air

1 3m / sv =

30 mB

A 2v

27 mD

C

Q

4m

3m P

90°

O

water

9 Mock Test-1

a. 1 2

1 2

NA NA

NA NA+ b. 1 2NA NA+

c. 1NA d. 2NA


In a thin rectangular metallic strip a constant current I flows

along the positive x-direction, as shown in the figure. The

length, width and thickness of the strip are ,ℓ w and d,

respectively. A uniform magnetic field B

is applied on the

strip along the positive y-direction. Due to this, the charge

carriers experience a net deflection along the z-direction.

This results in accumulation of charge carriers on the surface

PQRS and appearance of equal and opposite charges on the

face opposite to PQRS. A potential difference along the z-

direction is thus developed. Charge accumulation continues

until the magnetic force is balanced by the electric force. The

current is assumed to be uniformly distributed on the cross-

section of the strip and carried by electrons.

17. Consider two different metallic strips (1 and 2) of the

same material. Their lengths are the same, widths are

1w and 2w and thicknesses are

1d and 2 ,d respectively.

Two points K and M are symmetrically located on the

opposite faces parallel to the x-y plane (see figure). 1V

and 2V are the potential differences between K and M in

strips 1 and 2, respectively. Then, for a given current I

flowing through them in a given magnetic field strength

B, the correct statement(s) is(are)

a. If 1 2w w= and

1 22 ,d d= then 2 12V V=

b. If 1 2w w= and 1 22 ,d d= then 2 1V V=

c. If 1 22w w= and

1 2 ,d d= then 2 12V V=

d. If 1 22w w= and

1 2 ,d d= then 2 1V V=

18. Consider two different metallic strips (1 and 2) of same

dimensions (lengths ,ℓ width w and thickness d) with

carrier densities n1 and n2, respectively. Strip 1 is

placed in magnetic field B1 and strip 2 is placed in

magnetic field B2, both along positive y-directions.

Then V1 and V2 are the potential differences developed

between K and M in strips 1 and 2, respectively.

Assuming that the current I is the same for both the

strips, the correct option(s) is(are)

a. If 1 2B B= and

1 22 ,n n= then 2 12V V=

b. If 1 2B B= and

1 22 ,n n= then 2 1V V=

c. If 1 22B B= and

1 2 ,n n= then 2 10.5V V=

d. If 1 22B B= and

1 2 ,n n= then 2 1V V=

y

x

z

I

Q P

S

I W

K•

M•

R

d


l

Physics 10

JEE ADVANCE PAPER-II

Time 3 Hours. Max. Marks 180 (60 for Physics)

Read the Instructions Carefully



2. Section 1 contains 6 MCQs with one or more than one correct answer

Marking Scheme: +4 if only (all) the correct option(s) is (are) chosen, +3 if all the four options are correct but only three

options are chosen, +2 if three or more options are correct but only two options are chosen, both of the options must be

correct, +1 if two or more options are correct buy only one option is chosen and it must be correct, -2 (in all other cases)

3. Section 2 contains 8 Numerical value answer type questions.

Marking Scheme: +3 if only correct numerical value is given

4. Section 3 contains 4 Matching type questions with 4 options.

Marking Scheme: +3 if only the correct option is selected, -1 in all other cases

Note:

Possible response Section 1 Section 2 Section 3

Not attempted 0 0 0

Partial Correct +1 for each correct option selected 0 0

Correct +3 +4 +3

Incorrect -1 -2 0



1. A fission reaction is given by

236 140 94

92 54 38U Xe Sr ,x y→ + + +

where x and y are two

particles. Considering 236

92 U to at rest, the kinetic energies

of the products are denoted by Xe ,K

Sr ,K x (2MeV)K and

y (2MeV),K respectively. Let the binding energies per

nucleon of 236 140

92 54U, Xe and 94

38Sr be 7.5 MeV, 8.5 MeV

and 8.5 MeV respectively. Considering different

conservation laws, the correct option(s) is(are)

a. Sr Xe, , 129 MeV, 86 MeVx n y n K K= = = =

b. Sr Xe, , 129 MeV, 86 MeVx p y e K K

−= = = =

c. Sr Xe, , 129 MeV, 86 MeVx p y n K K= = = =

d. Sr Xe, , 86 MeV, 129 MeVx n y n K K= = = =

2. Two spheres P and Q of equal radii have densities 1ρ and

2 ,ρ respectively. The spheres are connected by a mass

less string and placed in liquids 1L and

2L of densities 1σ

and 2σ and viscosities

1η and 2 ,η respectively. They float

in equilibrium with the sphere P in 1L and sphere Q in 2L

and the string being taut (see figure). If sphere P alone in

2L has terminal velocity

PV and Q alone in 1L has

terminal velocity ,

QV then

a. 1

2

| |

| |

ηη

=

P

Q

V

V b. 2

1

| |

| |

ηη

=

P

Q

V

V

c. 0⋅ >

P QV V d. 0⋅ <

P QV V

3. In terms of potential difference V, electric current I,

permittivity 0 ,ε permeability 0µ and speed of light c, the

dimensionally correct equation(s) is(are)

a. 2 2

0 0µ ε=I V b. 0 0ε µ=I V

c. 0ε=I cV d.

0 0µ ε=cI V

4. Consider a uniform spherical charge distribution of radius

1R centred at the origin O. In this distribution, a spherical

1LP

Q 2L

11 Mock Test-1

cavity of radius R2, centred at P with distance

1 2= = −OP a R R (see figure) is made. If the electric field

inside the cavity at position r is ( ),

E r then the correct

statement(s) is(are)

a. E is uniform, its magnitude is independent of 2R but its

direction depends on r

b. E is uniform, its magnitude depends on

2R and its

direction depends on r

c. E is uniform, its magnitude is independent of a but its

direction depends on a

d. E is uniform and both its magnitude and direction

depend on a

5. In plotting stress versus strain curves for two materials P

and Q, a student by mistake puts strain on the y-axis and

stress on the x-axis as shown in the figure. Then the

correct statement(s) is(are)

a. P has more tensile strength than Q

b. P is more ductile than Q

c. P is more brittle than Q

d. The Young’s modulus of P is more than that of Q

6. A spherical body of radius R consists of a fluid of constant

density and is in equilibrium under its own gravity. If P(r)

is the pressure at r(r < R), then the correct option(s) is(are)

a. ( 0) 0= =P r b. ( 3 / 4) 63

( 2 / 3) 80

==

=

P r R

P r R

c. ( 3 / 5) 16

( 2 / 5) 21

==

=

P r R

P r R d.

( / 2) 20

( / 3) 27

==

=

P r R

P r R



7. An electron in an excited state of 2Li +

ion has angular

momentum 3 / 2 .πh The de Broglie wavelength of the

electron in this state is 0πp a (where

0a is the Bohr

radius). The value of p is

8. A large spherical mass M is fixed at one position and two

identical point masses m are kept on a line passing

through the centre of M (see figure). The point masses are

connected by a rigid mass less rod of length ℓ and this

assembly is free to move along the line connecting them.

All three masses interact only through their mutual

gravitational interaction. When the point mass nearer to M

is at a distance 3= ℓr from M, the tension in the rod is

zero for .288

Mm k

=

The value of k is

9. The energy of a system as a function of time t is given as 2( ) exp( α ),= −E t A t where 1α 0.2 .−= s The measurement

of A has an error of 1.25%. If the error in the

measurement of time is 1.50%, the percentage error in the

value of ( ) at 5=E t t s is

10. The densities of two solid spheres A and B of the same

radii R vary with radial distance r as ( )ρ =

A

rr k

R and

5

( ) ,ρ =

B

rr k

R respectively, where k is a constant. The

moments of inertia of the individual spheres about axes

passing through their centres are AI and ,BI respectively.

If ,10

=B

A

I n

Ithe value of n is

11. Four harmonic waves of equal frequencies and equal

intensities 0I have phase angles 0,

2,

3 3

π πand .π When

they are superposed, the intensity of the resulting wave is

0 .nI The value of n is

12. For a radioactive material, its activity A and rate of change

of its activity R are defined as = −dN

Adt

and ,= −dA

Rdt

where N(t) is the number of nuclei at time t. Two

radioactive sources P (mean lifeτ ) and Q (mean life 2 )τ

have the same activity at t = 0. Their rates of change of

activities at 2τ=t are PR and ,QR respectively. If

,=P

Q

R n

R e then the value of n is

M

r ℓ

m m

Str

ain

Stress

P

Q

1R

O

P R2

a

Physics 12

13. A monochromatic beam of light is incident at 60° on one

face of an equilateral prism of refractive index n and

emerges from the opposite face making an angle

( )nθ with the normal (see the figure). For 3=n the

value of θ is 60° and .d

mdn

θ= The value of m is

14. The figure shows a portion of an electric circuit. Resistors

are known and are indicated on the diagram and the

voltmeters are identical. If the voltmeters V1 and V2 read

7.5 V and 5.0 V respectively, find the reading of the volt-

meter V3.


Matching type questions with 4 options

15. Match the statement of Column with those in

Column II:

Column I Column II

(A) In any Bohr orbit of

the hydrogen atom,

the ratio of kinetic

energy to potential

energy of the electron

is

1. 1

2−

(B) The ratio of the

kinetic energy to the

total energy of an

electron in a Bohr

orbit is

2. 2

(C) In the lowest energy

level of hydrogen

atom, the electron has

the angular

momentum

3. 2

h

π

(D) Ratio of the

wavelengths of first

line of Lyman series

and first line of

Balmer series is

4. 5 : 27

5. 0

a. A→ 1; B→ 2, 4; C→3; D→ 4

b. A→3; B→1, 2; C→1; D→ 2

c. A→4; B→3, 4; C→2; D→ 1

d. A→2; B→2, 3; C→4; D→ 3

16. You are given many resistances, capacitors and inductors.

These are connected to a variable DC voltage source

(the first two circuits) or an AC voltage source of 50 Hz

frequency (the next three circuits) in different ways as

shown in Column II. When a current I (steady state for DC

or r.m.s. for AC) flows through the circuit, the

corresponding voltage V1 and V2. (indicated in circuits) are

related as shown in Column I.

Column I Column II

(A) 10,I V≠ is

proportional to I

1.

(B) 2 10,I V V≠ > 2.

(C) 1 20,V V V= = 3.

(D) 20,I V≠ is

proportional to I

4.

5.

a. A→3, 4; B→2, 3, 4; C→1, 2; D→2, 3, 4

b. A→1, 2; B→1, 2, 3; C→3, 4; D→1, 3 ,4

c. A→2, 4; B→2, 3, 4; C→2, 3; D→1, 2, 3

d. A→1, 3; B→1, 3, 4; C→1, 3; D→2, 3, 4

V1 V2

1 kΩ

V

3µF

V1 V2

6 mH

V

3µF

V1 V2

6 mH

V

2 Ω

V1 V2

6 mH

V

2 Ω

V1 V2

6 mH

V

3 µF

R1 = 9

R2 = 14

R3 = 24 V3

V2

V1

θ 60°

13 Mock Test-1

17. The gravitational field intensity E

of earth at any point is

defined as the gravitational force per unit mass at that

point. If varies from place to place. The variation is shown

in column II with form of position r

v s the E

in the form

of graphs. The variation of r

is given in column I.

Choose the correct form of graphs for the corresponding

variations of .r

Column I Column II

(A) Position r

of body

measured from

surface earth

upward

1.

(B) Position r

of body

measured from

surface of earth

along diameter to

opposite point on

surface of earth

2.

(C) Position r

measured

from center of earth

to any point

3.

(D) Position r

measured

from center of

hollow sphere to

any point

4.

a. A→1; B→2; C→3; D→4

b. A→4; B→3; C→2; D→1

c. A→4; B→1; C→3; D→2

d. A→3; B→2; C→1; D→4

18. A person in a lift is holding a water jar, which has a small

hole at the lower end of its side. When the lift is at rest, the

water jet coming out of the hole hits the floor of the lift at a

distance d of 1.2 m from the person. In the following, state

of the lift’s motion is given in Column I and the distance

where the water jet hits the floor of the lift is given in

Column II. Match the statements from Column I with those

in Column II and select the correct using the code given

below the lists.

Column I Column II

(A) Lift is accelerating

vertically up.

1. d = 1.2 m

(B) Lift is accelerating

vertically down with

an acceleration less

than the gravitational

acceleration.

2. d > 1.2 m

(C) Lift is moving

vertically up with

constant speed.

3. d < 1.2 m

(D) Lift is falling freely. 4. No water leaks out of

the jar

a. A→1; B→1; C→1; D→4

b. A→4; B→3; C→2; D→1

c. A→4; B→1; C→3; D→2

d. A→3; B→2; C→1; D→4

O R

E

rE

O

E

r

O R

E

rE

O R

E

rE


Physics 14

ANSWER & SOLUTIONS

JEE-Main

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

c a c b c b c b b b

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

d b b a c d d c d b

21. 22. 23. 24. 25.

a b c c c

1. (c) Displacement vector ˆˆ ˆr xi yj zk= ∆ + ∆ + ∆

ˆˆ ˆ ˆ ˆ(3 2) (4 3) (5 5)i j k i j= − + − + − = +

2. (a) Tangential force (Ft) of the bead

will be given by the normal reaction

(N), while centripetal force (Fc) is

provided by friction (Fr). The bead

starts sliding when the centripetal

force is just equal to the limiting

friction. Therefore, tF ma m L Nα= = =

∴ Limiting value of frictionmax( )rf N m Lµ µ α= = . . . (i)

Angular velocity at time t is tω α=

∴ Centripetal force at time t will be

2 2 2 cF mL mL tω α= = . . . (ii)

Equation (i) and (ii), we get tµα

=

For ( )max

, . .c rt F f i eµα

> > , the bead starts sliding.

In the figure Ft is perpendicular to the paper inwards.

3. (c) Power F v Fv= = i

( ) volume

ddm

F v vdt dt

ρ × = =

( )volumedv

dtρ

=

( ) 2v Av Avρ ρ= =

∴ Power 3P Avρ= or 3P v∝

4. (b) Let ρ be the density of lead. Then 34

3M Rπ ρ= =

= mass of total sphere

3

1

4

3 2

Rm π ρ = =

mass of removed part

8

M =

2

7

8 8

M Mm M= − = = mass of remaining sphere

Choosing the centre of big sphere as the origin,

1 1 2 2

1 2

CM

m x m xX

m m

+=

+

( ) ( ) ( ) 2/8 / 2 7 /8

0M R M x

M

× + ×=

Solving, we get 214

Rx

−=

5. (c) In air effg g=

In water, Buoyant force

Masseffg = g –

4

w

b

d gg g

d= − =

Now 2 ,eff

tt

gπ= Hence 02t t=

6. (b) Soap solution has lower surface tension as compared

to pure water so h is less for soap solution.

7. (c) 1 2F F=

⇒ 1 2Y A Y A1 2α ∆θ = α ∆θ ⇒ 1 2

2 1

3

2

Y r

Y r= =

8. (b) Let 0°C be the reference temperature for zero heat,

then initial heat energy = final heat energy

1 1 1 2 2 2 3 3 3m s m s m sθ θ θ+ + ( )1 1 2 2 3 3m s m s m s= + +

⇒ θ 1 1 1 2 2 2 3 3 3

1 1 2 2 3 3

m s m s m s

m s m s m s

θ θ θθ

+ +=

+ +

1 1 1 2 2 2 3 3 3

1 1 2 2 3 3

Vd s Vd s Vd s

Vd s Vd s Vd s

θ θ θ+ +=

+ +1 1 1 2 2 2 3 3 3

1 1 2 2 3 3

d s d s d s

d s d s d s

θ θ θ+ +=

+ +

9. (b) Speed of sound in a gas is given by1

,RT

v vM M

γ= ∝

∴ 1 2 2

2 1 1

v M m

v M m= =

Here, γ 5

3

p

V

C

Cγ = = for both the gases

monoatomic

5

3γ =

10. (b)

1

1 1 2 11 1 2 2

1 2

V TT V T V

V T

γ

γ γ

−

− − = ⇒ =

or

1 1

1 1 1

2 2 22

V T T

V T T

γ γ− −

= = ( )1

1 1

22γ−= >

11. (d) Wien’s displacement law is mT bλ = (b = Wien’s

constant)

∴ 62.88 10 nm-K

2880Km

b

Tλ

×= =

y

x

60o

F sin 60o

F c

os

60

o

F

A L

cF

tF

is inwardstF

15 Mock Test-1

∴ λ = 1000 nm

Energy distribution with wavelength will be as follows:

From the fig. it is clear that 1

2

U

U (In fact U2

is maximum)

12. (b) Work done by the field

( ) ( )B A C Ae V V e V V= − = − (∵ VB = VC)

⇒ 19

19

6.4 10( ) 4

1.6 10C A

WV V V

e

−

−

×− = = =

× 4V

13. (b) In the electric field between plates of parallel plate

capacitor 0

Eσε

=

∴ 0

qF qE

σε

= = . . . (i)

When one plate is removed, the electric field becomes

02

Eσε

′ =

∴ 02

F qEσε

′ ′= = . . . (ii)

1

2 2

F FF

F

′′= ⇒ =

14. (a) Resistance of 9 mm cable 5= Ω

As 1

RA

∝ or2

1;R

r∝ resistance of 3 mm

Cable 9 5 45× = Ω

In second case 6 wires are connected in parallel, so total

resistance of cables 45

7 56

= = ⋅ Ω

15. (c) Magnetic field at any point lying on the current

carrying straight conductor is zero. Here

H1 = Magnetic field at M due to current in PQ.

H2 = Magnetic field at M due to QR + magnetic field at M

due to QS + magnetic field at M due to PQ

11 1

30

2 2

HH H= + + = ⇒ 1

2

2

3

H

H=

16. (d) If monochromatic light is replaced by white light, a

few coloured fringes are seen on either side of a central

white fringe.

17. (d) Electric field of an electromagnetic wave in free space

is given by 7 ˆ10cos(10 )E t kx j= +

which is acting along

y-direction. As E is varying with x and t, hence

propagation of electromagnetic wave takes place along –x

axis. Thus statement (d) is wrong. Comparing the relation,

710cos(10 )E t kx= + with standard equation of

electromagnetic wave

0 0

2 2 2cos ( ) cosE E t x E t x

π πυ πυ

λ λ λ = + = +

We have, 0 10 / .E V m= Thus statement (c) is correct.

7210

πυλ

= or 8

72 (3 10 )10

πλ

× ×=

or 22

60 60 188.47

λ π= = × ≈ m

Thus, statement (a) is correct.

⇒ 2 2 1

0.033 /60 30

k rad mπ πλ π

= = = = rad / m

Thus, statement (b) is wrong.

18. (c)hc

E Wλ

= = − . . . (i)

2hc

E Wλ

= = −′

. . . (ii)

Dividing (ii) by (i)2

2 2

hcW

hc hcW W

hcW

λλ λ

λ

−′= = − = −

′−

⇒ 2hc hc

Wλ λ

= −′

⇒ 1 2 W

hcλ λ= −

′

⇒ 1

2 W

hc

λ

λ

′ =−

⇒ 2

λλ′ > but less than λ

19. (d) Energy required to remove electron in the n = 2

state2

13.63.4

(2)eV= + = + eV

20. (b) Hydrogen bomb is based on nuclear fusion.

21. (a) 3 2 3(1.2 10 m)V l

−= = × 6 31.728 10 m−= ×

∵ Length (l) has two significant figures, the volume (V) will

also have two significant figures. Therefore, the correct

answer is 6 31.7 10 mV lV −= = ×

22. (b) sin ,a g lθ= = sin

h

θ

⇒ 21sin

sin 2

hg t= θ

θ⇒ t ∝

1

sin θ

⇒ 2 1

1 2

sin

sin

t

t=

θ

θ2 1

1 2

sin 1/ 2 1

sin 3 / 2 3= = = 1

2

3/ 3

3 3

tt = = = s

499

5

99

999

1

000

1499

1

500

( ) ( )A BW q dV e V V= − = − −

Physics 16

23. (c)

( )

1 2

221

GM GM

x x=

−

⇒ ( )22

100 10, 000

1x x=

−

or ( )

2

2

1

100 1

x

x=

−⇒

21 1

10 1 11

xx

x= ⇒ =

−m

24. (c) In C.G.S. 3

29axial

MB

x= = . . . (i)

equaterial 3 3

8

2

M MB

xx= =

. . . (ii)

From equation (i) and (ii) equaterial 36B = gauss.

25. (c) rvn n nr

′ = +/ 2 3

2

vn n n

v= + =

% change in frequency

100% 1 100%n n n

n n

′ ′− = × = − ×

31 100% 50%

2

= − × =

JEE Advance Paper -I

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

c a,b,c a,c,d b,d a,c,d a,b,d 9 6 6 5

11. 12. 13. 14. 15. 16. 17. 18.

7 2 8 5 a,c d a,d a,c

1. (c) For air to glass 1

1.5 1.4 1 1.5 1.4− −= +

f R R

∴∴∴∴ 1 3=f R

For glass to air. 2

1 1.4 1.5 1 1.4− −= +

− −f R R

∴∴∴∴ 2 2=f R

2. (a, b, c) ˆ2 ( )[ ]F I L R i B= + ×

3. (a, c, d) 2 1 1 1 3 2[ ], [ ], [ ]h ML T c LT G M L T

− − − −≡ ≡ ≡

⇒ 3

,hc hG

M LG c

∝ ∝

4. (b, d) For first oscillator 2 2

1 1

1

2E m aω=

and 1p mv m a bω= = =

⇒ 1

1a

b mω= . . .(i)

For second oscillator

2 2

2 2 2

1, and 1

2E m R mω ω= =

. . .(ii)

⇒ 22

1

an

b

ωω

= =

⇒ 1 2

2 2 2 2

1 2

E E

a aω ω=

⇒ 1 2

1 2

E E

ω ω=

5. (a, c, d) Since, the spool rolls over the horizontal surface,

therefore, instantaneous axis of rotation passes through the

point of contact of spool with the horizontal surface.

About the instantaneous axis of rotation, moment

produced by F is clockwise. Therefore, the spool rotates

clockwise. In that case acceleration will be rightward and

thread will wind. If rotational motion of spool is

considered about its own axis then resultant moment on it

must be clockwise. But moment produced by the force F

is anticlockwise and its magnitude is equal to F.r,

Hence, moment produced by the friction (about its own

axis) must be clockwise and its magnitude must be greater

than F.r. It is possible only when friction acts leftwards.

Therefore, option (b) is correct.

6. (a, b, d) 1 1 21

1 3

( )+=

+

R V VV

R R⇒

1 3 2 1=V R V R

⇒ 3 1 22

1 3

( )+=

+

R V VV

R R⇒

2 1 2 3=V R V R

7. (9) For refraction at the first surface, 8 cm,u = −

1 1 28 cm, 1, 1.5R µ µ= − = =

⇒ 2 1 2 1

1v u R

µ µ µ µ−− =

⇒ 1.5 1 0.5

' 8 8v+ =

−⇒ ' 8 cmv = −

It means due to the first surface the image is formed at the

centre.

For the second surface 1 29 cm, 1.5, 9 cmu Rµ= − = = −

⇒ 2 1 2 1

2

m m m m

v u R

−− =

⇒ 1 1.9 1 1.5

5 9v

−+ =

−⇒ 9 cmv = −

Thus, the final image is formed at the centre of the sphere.

8. (6) From the figure θ = 60°

So, No. of rectangular surfaces used to form a360

660

°= =

°m

θ3

2a

Line change

Rectangular

surface a

2(L + R)

1 r1 P r2

Planet 2

17 Mock Test-1

9. (6) As is clear from figure. 1 2 3E E E= +

1 2 3

hc hc hc

λ λ λ= +

∴∴∴∴ 2 1 3

1 1 1 1 1 1

2 3 6λ λ λ= − = − = = 6 units

10. (5) Using work energy theorem + = ∆mg FW W KE

− + = ∆mgh Fd KE 1 10 4 18(5)− × × + = ∆KE

⇒ 50∆ =KE ∴∴∴∴ 5=n

11. (7) Kinetic energy of a pure rolling disc having velocity of

centre of mas 2 2

2 2

2

1 1 3

2 2 2 4

mR vv mv mv

R

= + =

So, 2 2

2

3 3(3) (30) ( ) (27)

4 4m mg m v mg+ = +

∴∴∴∴ 2 7 m / sv =

12. (2) 410

A B

dQ dQ

dt dt

=

2 4 4 2 4(400 ) 10 ( )A BR T R T=

So, 2 and 2A B

A B

B A

TT T

T

λλ

= = =

13. (8) Here, 2h, ?T t= =

To work safely with the sample, its activity must be

reduced to 1

.16

From

4

0

1 1 1

2 16 2

nN

N

= = =

∴∴∴∴ n = 4

⇒ 2 4 8ht n T= = × =

14. (3) For maxima, 2 2 2 24,

3d x d x mλ+ − + = m is an

integer

So, 2 2 2 29x m dλ= − ⇒ 3=p

15. (a, c) cθ ≥

⇒ 90°− ≥r c

⇒ sin(90 )° − ≥r c

⇒ cos sin≥r c

Using sin

sin

i

m

ni

r n= and 2

3

sin =n

n

We get, 2 2

2 1 2

2sin

−=m

m

n ni

n

Putting values, we get, correct options as a & c

16. (d) For total internal reflection to take place in both

structures, the numerical aperture should be the least one

for the combined structure and hence, correct option is d.

17. (a, d) 1 2

=I I

⇒ 1 1 2 2=neA v neA v

⇒ 1 1 1 2 2 2=d w v d w v

Now, potential difference developed across MK, =V Bvw

⇒ 1 1 1 2

2 2 2 1

= =V v w d

V v w d and hence correct choice is a & d.

18. (a, c) As 1 2=I I

⇒ 1 1 1 1 2 2 2 2=n w d v n w d v

Now, 2 2 2 2 2 2 1 1 1 2 1

1 2 1 1 1 1 2 2 2 1 2

= = =

V B v w B w n wd B n

V B v w B w n w d B n

JEE Advance Paper -II

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

a a,d a,c d a,b b,c 2 7 4 6

11. 12. 13. 14. 15. 16. 17. 18.

3 2 2 3 a a d a

1. (a) Q value of reaction

(140 94) 8.5 236 7.5 219= + × − × = MeV

So, Total kinetic energy of Xe and Sr 219 2 2 215= − − = MeV

So, by conservation of momentum, energy, mass and

charge, only option a. is correct.

2. (a, d) From the given conditions, 1 1 2 2

ρ σ σ ρ< < <

From equilibrium, 1 2 1 2σ σ ρ ρ+ < = +

1

2

2

9

ρ ση

2 −=

pV g and 2 1

1

2

9

ρ ση

−=

QV g

So, 1

1

| |

| |

P

Q

V

V

ηη

= and . 0P QV V

<

3. (a, c) BI c VI≡ℓ

⇒ 2

0I c VIµ ≡

⇒ 0Ic Vµ =

⇒ 2 2 2 2

0µ =I c V

⇒ 2 2

0 0I Vµ ε=

⇒ 0cV Iε =

4. (d) 1 2

03

ρε

=

E CC

C1 ⇒ Centre of sphere and

C2 ⇒ centre of cavity.

5. (a, b) stress

strain=Y

⇒ 1 strain

stress=

Y

1nmn

i

r

θ

n2

ρ

Physics 18

⇒ 1 1

P QY Y> ⇒ P QY Y<

6. (b, c)2

2( ) 1

= −

rP r K

R

7. (2)3

2 2π π= =nh h

mvr

de-Broglie Wavelength h

mvλ =

2

00

(3)2 22

3 3 Li

ara

z

π ππ= = =

8. (7) For m closer to M; 2

2 29− =

ℓ ℓ

GMm Gmma . . .(i)

and for the other m: 2

2 216+ =

ℓ ℓ

Gm GMmma . . .(ii)

From both the equations, 7=k

9. (4) 2( )

−= atE T A e

⇒ 2 2α − −= − +at atdE A e dt AdAe

Putting the values for maximum error,

⇒ 4

100=

dE

E ⇒ % error = 3.

10. (6) 2 224

3ρ π= ∫I r r dr

⇒ 2 2( )( )( )∝ ∫AI r r r dr

⇒ 5 2 2( )( )( )∝ ∫BI r r r dr

∴

6

10=B

A

I

I

11. (3) First and fourth wave interfere destructively. So from

the interference of 2nd

and 3rd

wave only,

⇒ 0 0 0 0 0

22 cos 3

3 3

π π = + + − =

netI I I I I I ⇒ 3=n

12. (2) 1 1

;2

P Qλ λτ τ

= = ⇒ 0

0

( ) p

Q

t

PP

t

Q Q

A eR

R A e

λ

λ

λ

λ

−

−=

At 2 ;τ=t 2P

Q

R

R e=

13. (2) Snell’s Law on 1st surface: 1

3sin

2= n r

1

3sin

2=r

n . . .(i)

⇒ 2

1 2

3 4 3cos 1

4 2

−= − =

nr

n n

1 2

60+ = °r r . . .(ii)

Snell’s Law on 2nd

surface:2sin sinn r = θ

Using equation (i) and (ii) 1sin(60 ) sinθ° − =n r

⇒ 1 1

3 1cos sin sin

2 2n r r

− = θ

⇒ ( )234 3 1 cos

4

d dn

dn dn

θ− − = θ

For 60θ = ° and 3=n ⇒ 2d

dn

θ=

14. (3) ( )

( ) ( )1 2 2 1

3

1 3 1 2 3 2

3.0VVV R R

VV R R V R R

−= =

− − −

15. (a) A→ 1; B→ 2, 4; C→ 3; D→ 4

16. (a) A→ 3, 4; B→ 2, 3, 4; C→ 1, 2; D→ 2, 3, 4

17. (d) Gravitational field intensity at any point outside due to

earth is given by 2 2

( )

GM GME g

R h r= = =

+

where h is height from surface or 2

1E

r∝

but – ve in sign.

Graph is parabolic.

Inside earth g decreases are r increases form surface of

earth.

1x g

E g g xR R

= + = −

2

1E

r− ∝

or E x∝ is straight line decreasing to 0 at centre of earth and

then increases in magnitude but – ve as directed towards

the centre always.

At any point outside 2

1E

r∝ (parabolic graph)

Inside earth E r∝ (linear graph)

In hollow sphere inside 0E = graph along r-axis; outside

as if whole mass lies at centre 2

1E

r∝ decreasing

parabolic graph after sudden rise at surface of earth.

18. (a) A→1; B→1; C→1; D→4

In A, B, C no horizontal velocity is imparted to falling

water, so d remains same.

In D, since its free fall, 0effa =

∴ Liquid won’t fall with respect to lift.

19 Mock Test-2



1. In the relation Z

kp eα

−θ

α=

β p is pressure, Z is distance, k

is Boltzmann constant and θ is the temperature. The

dimensional formula of α will be:

a. 2[ ]MLT− b. 2[ ]ML T

c. 0 1[ ]ML T− d. 0 2 1[ ]M L T

−

2. A small block is shot into each of the four tracks as shown

below. Each of the tracks rises to the same height. The

speed with which the block enters the track is the same in

all cases. At the highest point of the track, the normal

reaction is maximum in

a. b.

c. d.

3. A particle, which is constrained to move along x-axis, is

subjected to a force in the same direction which varies

with the distance x of the particle from the origin as 3( ) .F x kx ax= − + Here, k and a are positive constants.

For 0,x ≥ the functional form of the potential energy

U(x) of the particle is

a. b.

c. d.

4. A carpenter has constructed a toy as

shown in the adjoining figure. If the

density of the material of the sphere

is 12 times that of cone, the position

of the center of mass of the toy is

given by.

a. at a distance of 2R from O

b. at a distance of 3R from O

c. at a distance of 4R from O

d. at a distance of 5R from O

5. The total energy of a particle executing simple harmonic

motion is:

a. ∝ x b. ∝ x2

c. independent x d. ∝ x1/2

6. Two satellites A and B go around a planet P in circular

orbits having radii 4R and R respectively. If the speed of

satellite A is 3v, the speed of satellite B will be:

a. 12 v b. 6 v

c. 4

3v d.

3

2v

7. The following diagram shows three soap bubbles A, B and

C prepared by blowing the capillary tube fitted with stop

cocks, S1, S2 and S3. With stop cock S closed and stop

cocks S1, S2 and S3 opened.

a. B will start collapsing with volumes of A and C

increasing

b. C will start collapsing with volumes of A and B

increasing

c. C and A both will start collapsing with the volume of B

increasing

d. Volumes of A, B and C will becomes equal at

equilibrium

8. A steel wire of length 20 cm and uniform cross-section

1 mm2 is tied rigidly at both the ends. The temperature of

the wire is altered from 40°C to 20°C. What is the

magnitude of force developed in the wire? (Coefficient of

linear expansion for steel, α = 1.1 × 10–5

/°C and Y for

steel is 11 22.0 10 N/m× ).

a. 62.2 10× N b. 16 N

c. 8 N d. 44 N

9. The intensity level of two waves of same frequency in a

given medium are 20 dB and 60 dB. Then the ratio of

their amplitudes is:

a. 1 : 4 b. 1 : 16

c. 1 : 410 d. 1 : 100

10. The specific heat of a substance varies as (2t2 + t) ×10

–3

cal/g °C. What is the amount of heat required to raise the

temperature of 100 g of substance through 20°C to 40°C.

S3 S1

S S2

B A

C

U(x)

x

U(x)

x

U(x)

x

U(x)

x

v v

v v

O2 2R

2R

O1

O

2R

Physics 20

a. 37.9 kilocal b. 3.79 kilocal

c. 379 kilocal d. 82 cal

11. A real gas behaves like an ideal gas if its

a. pressure and temperature are both high

b. pressure and temperature are both low

c. pressure is high and temperature is low

d. pressure is low and temperature is high

12. The plates of a parallel plate capacitor of capacity of 50µF

are charged by a battery to a potential of 100 volt. The

battery remains connected the plates are separated from

each other so that the distance between them is doubled.

How much is the energy spent by battery in doing so?

a. 25 × 10–2

J b. –12.5 × 10–2

J

c. – 25 × 10–2

J d. 12.5 × 10

–2 J

13. The magnetic moment produced in a substance of 1gm is

6 × 10–7

ampere-metre2. If its density is 5 gm/cm

3, then

the intensity of magnetisation in A/m will be

a. 8.3 × 106 b. 3.0

c. 1.2 × 10–7

d. 3 × 10–6

14. An ionized gas contains both positive and negative ions. If

it is subjected simultaneously to an electric field along the

+x direction and a magnetic field along the +z direction,

then

a. Positive ions deflect towards +y direction and negative

ions towards –y direction

b. All ions deflect towards +y direction

c. All ions deflect towards –y direction

d. Positive ions deflect towards –y direction and negative

ions towards +y direction

15. A hundred turns of insulated copper wire are wrapped

around an iron cylinder of area 1 × 10–3

m2 are connected

to a resistor. The total resistance in the circuit is 10 ohms.

If the longitudinal magnetic induction in the iron changes

from 1 Wb m–2

, in one direction to 1 Wb 2m− in the

opposite direction, how much charge flows through the

circuit?

a. 2 × 10–2

C b. 2 × 10–3

C

c. 2 × 10–4

C d. 2 × 10–5

C

16. A resistor R, an inductor L and a capacitor C are

connected in series to an oscillator of frequency n. If the

resonant frequency is nr, then the current lags behind

voltage, when

a. n = 0 b. n < nr

c. n = nr d. n > nr

17. The intensity ratio at a point of observation due to two

coherent waves is 100 : 1. The ratio between their

amplitudes is:

a. 1 : 1 b. 1 : 10

c. 1 : 100 d. 10 : 1

18. An electromagnetic wave propagating along north has its

electric field vector upwards. Its magnetic field vector

points towards

a. north b. east

c. west d. downwards

19. The ionisation energy of 10 times ionised sodium atom is

a. 13.6 eV b. 13.6 11× eV

c. 13.6

11 eV d. 213.6 (11)× eV

20. The example of nuclear fusion is

a. Formation of barium and krypton from uranium

b. Formation of helium from hydrogen

c. Formation of plutonium 235 from uranium 235

d. Formation of water from hydrogen and oxygen


21. A force of 5 N acts on a particle along a direction making

an angle of 60° with vertical. Its vertical component be:

a. 10 N b. 3 N

c. 4 N d. 2.5 N

22. A neutral water molecule (H2O) in it's vapor state has an

electric dipole moment of magnitude 6.4 ×10–30

C–m.

How far apart are the molecules centres of positive and

negative charge?

a. 4 m b. 4 mm

c. 4 µm d. 4 pm

23. A cell of e.m.f. 1.5 V having a finite internal resistance is

connected to a load resistance of 2 .Ω . For maximum

power transfer, the internal resistance of the cell in ohms

should be:

a. 4 b. 0.5

c. 2 d. None of these

24. Two plane mirrors are at 45° to each other. If an object is

placed between them, then the number of images will be

a. 5 b. 9

c. 7 d. 8

25. A photon of energy 8 eV is incident on a metal surface of

Threshold frequency 1.6×1015

Hz. The kinetic energy of the

photoelectrons emitted (in eV) 34(Take 6 10 J s)h −= × −

a. 1.6 b. 6

c. 2 d. 1.2

21 Mock Test-2

JEE ADVANCE PAPER-I



1. A piece of wire is bent in the shape of a parabola 2=y kx

(y-axis vertical) with a bead of mass m on it. The bead can

slide on the wire without friction. It stays at the lowest

point of the parabola when the wire is at rest. The wire is

now accelerated parallel to the x-axis with a constant

acceleration a. The distance of the new equilibrium

position of the bead, where the bead can stays at rest with

respect to the wire, from the y-axis is

a. a

gk b.

2

a

gk c.

2a

gk d.

4

a

gk

2. A block of mass m is placed on a surface with a vertical

cross-section given by

3

.6

xy = If the coefficient of friction

is 0.5, the maximum height above the ground at which the

block can be placed without slipping is:

a. 1

m3

b. 1

m2

c. 1

m6

d. 2

m3

3. A ball moves over a fixed track as shown in the figure.

From A to B the ball rolls without slipping. Surface BC is

frictionless. KA, KB and KC are kinetic energies of the ball

at A, B and C, respectively. Then

a. ;> >A C B Ch h K K b. ;> >A C C Ah h K K

c. ;= =A C B Ah h K K d. ;< >A C B Ch h K K

4. A small block of mass of 0.1 kg lies on a fixed inclined

plane PQ which makes an angle θ with the horizontal. A

horizontal force of 1 N acts on the block through its centre

of mass as shown in the figure. The block remains

stationary if (take g = 10 m/s2).

a. 45θ = °

b. 45θ > ° and a frictional force acts on the block towards P.

c. 45θ < ° and a frictional force acts on the block towards

d. a frictional force acts on the block towards

5. In the figure, a ladder of mass m is shown leaning against

a wall. It is in static equilibrium making an angle θwith

the horizontal floor. The coefficient of friction between

the wall and the ladder is µ1 and that between the floor

and the ladder is µ2. The normal reaction of the wall on

the ladder is N1 and that of the floor is N2.

If the ladder is

about to slip, then

a. 1 20, 0= ≠µ µ

and

2 tan2

mgN θ =

b. 1 20, 0≠ =µ µ

and

1tan

2

mgN θ =

c. 1 20, 0≠ ≠µ µ and

1

1 21

mgN =

+ µ µ

d. 1 20, 0= ≠µ µ

and 1 tan

2

mgN θ =

6. Two small particles of equal masses start moving in

opposite directions from a point A in a horizontal circular

orbit. Their tangential velocities are v and 2v

respectively, as shown in the figure. Between collisions,

the particles move with constant speeds. After making

how many elastic collisions, other than that at A, these

two particles will again reach the point A?

a. 4 b. 3

c. 2 d. 1

A 2v v

1µ

2µ

θ

O

Q

1 N

P θ

hc hA

A

B

C

µ1

µ2

Physics 22



7. The densities of two solid spheres A and B of the same

radii R vary with radial distance r as ( )ρ =

A

rr k

R and

5

( ) ,ρ =

B

rr k

R respectively, where k is a constant.

The moments of inertia of the individual spheres about

axes passing through their centres are IA and IB,

respectively. If ,10

=B

A

I n

Ithe value of n is

8. Gravitational acceleration on the surface of a planet is

6,

11g where g is the gravitational acceleration on the

surface of the earth. The average mass density of the

planet is 1

3times that of the earth. If the escape speed on

the surface of the earth is taken to be 11 kms–1

, the escape

speed on the surface of the planet in kms–1

will be

9. A bullet is fired vertically upwards with velocity v from

the surface of a spherical planet. When it reaches its

maximum height, its acceleration due to the planet’s

gravity is th1/ 4 of its value at the surface of the planet. If

the escape velocity from the planet is ,escv v N= then the

value of N is (ignore energy loss due to atmosphere)

10. The ends Q and R of two thin wires, PQ and RS, are

soldered (joined) together. Initially each of the wires has a

length of 1m at 10 C.° Now the end P is maintained at

10 C,° while the end S is heated and maintained at

400 C.° The system is thermally insulated from its

surroundings. If the thermal conductivity of wire PQ is

twice that of the wire RS and the coefficient of linear

thermal expansion of PQ is 5 11.2 10 ,K− −× the change in

length of the wire PQ is:

11. Two soap bubbles A and B are kept in a closed chamber

where the air is maintained at pressure 8 N/m2. The radii

of bubbles A and B are 2 cm and 4 cm, respectively.

Surface tension of the soap-water used to make bubbles is

0.04 N/m. Find the ratio ,B

A

n

n where An and Bn are the

number of moles of air in bubbles A and B, respectively.

[Neglect the effect of gravity.]

12. A cylindrical vessel of height 500 mm has an orifice

(small hole) at its bottom. The orifice is initially closed

and water is filled in it up to height H. Now the top is

completely sealed with a cap and the orifice at the bottom

is opened. Some water comes out from the orifice and the

water level in the vessel becomes steady with height of

water column being 200 mm. Find the fall in height

(in mm) of water level due to opening of the orifice.

[Take atmospheric pressure 5 21.0 10 N / m ,= × density of

water = 1000 kg/m3 and g = 10 m/s

2. Neglect any effect of

surface tension.]

13. Consider two solid spheres P and Q each of density 8 gm

3cm−and diameters 1 cm and 0.5 cm, respectively. Sphere

P is dropped into a liquid of density 0.8 gm 3cm− and

viscosity 3η = poiseulles. Sphere Q is dropped into a

liquid of density 1.6 gm 3cm− and viscosity 2η =

poiseulles. The ratio of the terminal velocities of P and Q

is:

14. A 0.1 kg mass is suspended from a wire of negligible

mass. The length of the wire is 1m and its cross-sectional

area is 7 24.9 10 m .−× If the mass is pulled a little in the

vertically downward direction and released, it performs

simple harmonic motion of angular frequency 140 rad 1.s− If the Young’s modulus of the material of the wire is

9 110 Nm ,n −× the value of n is


Paragraph based questions (2 paragraphs, each having 2

MCQs with one correct answer only)


A small block of mass 1 kg is released from rest at the top of a

rough track. The track is circular arc of radius 40 m. The block

slides along the track without toppling and a frictional force

acts on it in the direction opposite to the instantaneous velocity.

The work done in overcoming the friction upto the point Q, as

shown in the figure, below, is 150 J. (Take the acceleration due

to gravity, 210 m / s ).g −=

R

P R

Q

30°

23 Mock Test-2

15. The speed of the block when it reaches the point Q is

a. 15 ms− b. 110 ms−

c. 110 3 ms −

d. 120 ms−

16. The magnitude of the normal reaction that acts on the

block at the point Q is

a. 7.5 N

b. 8.6 N

c. 11.5 N

d. 22.5 N


Two waves 1 cos(0.5 100 )y A x tπ π= −

2 cosy A=

(0.4 92 )x tπ π− are travelling in a pipe placed along x-axis.

17. Find the number of times intensity is maximum in time

interval of 1 sec

a. 4 b. 6

c. 8 d. 10

18. Find wave velocity of louder sound

a. 100 m/s b. 192 m/s

c. 200 m/s d. 96 m/s


Physics 24




1. A person trying to lose weight by burning fat lifts a mass

of 10 kg upto a height of 1 m 1000 times. Assume that the

potential energy lost each time he lowers the mass is

dissipated. How much fat will he use up considering the

work done only when the weight is lifted up? Fat supplies 73.8 10 J× of energy per kg which is converted to

mechanical energy with a 20% efficiency rate (Take 2

9.8 ms ) :g−

=

a. –32.45 10 kg× b. –3

6.45 10 kg×

c. –39.89 10 kg× d. –3

12.89 10 kg×

2. If the resultant of all the external forces acting on a system

of particles is zero, then from an inertial frame, one can

surely say that

a. linear momentum of the system does not change in

time

b. kinetic energy of the system does not change in time

c. angular momentum of the system does not change in

time

d. potential energy of the system does not change in time

3. A point mass of 1 kg collides elastically with a stationary

point mass of 5 kg. After their collision, the 1 kg mass

reverses its direction and moves with a speed of 2

1ms .−Which of the following statement(s) is (are) correct

for the system of these two masses?

a. Total momentum of the system is 3 kg 1ms−

b. Momentum of 5 kg mass after collision is 4 kg 1ms−

c. Kinetic energy of the centre of mass is 0.75 J

d. Total kinetic energy of the system is 4 J

4. A solid sphere of mass M, radius R and having moment of

inertia about an axis passing through the centre of mass as

I, is recast into a disc of thickness t, whose moment of

inertia about an axis passing through its edge and

perpendicular to its plane remains I. Then, radius of the

disc will be

a. 2

15

R b.

2

15R

c. 4

15

R d.

4

R

5. A ball of mass (m) 0.5 kg is attached to the end of a string

having length (L) 0.5 m. The ball is rotated on a

horizontal circular path about vertical axis. The maximum

tension that the string can bear is 324 N. The maximum

possible value of angular velocity of ball (in rad/s)

is

a. 9 b. 18

c. 27 d. 36

6. A small mass m is attached to a massless string whose

other end is fixed at P as shown in the figure. The mass is

undergoing circular motion in the x-y plane with centre at

O and constant angular speed ω . If the angular

momentum of the system, calculated about O and P are

denoted by

OL and P

L

respectively, then.

a.

OL and

PL do not vary with time

b.

OL varies with time while

PL remains constant.

c.

OL remains constant while

PL varies with time

d.

OL and

PL both vary with time.



7. A body sends waves 500 mm long through medium A and

0.25 m long in medium B. If velocity of waves in medium

A is 16 m/s, what is the velocity (in m/s) of waves in

medium B?

8. A light pointer fixed to one prong of a tuning fork touches

gently a smoked vertical plate. The fork is set vibrating

and the plate is allowed to fall freely. Two complete

oscillations are traced when the plate falls through 40 cm.

What is the frequency (in Hz) of the tuning fork?

9. A water tank is 20 m deep. If the water barometer reads

10 m at that place, then what is the pressure at the bottom

of the tank in atmosphere?

m

P

z

O

L

m

25 Mock Test-2

10. There is a soap bubble of radius 42.4 10

−× m in air cylinder

which is originally at the pressure of 5 210 N/m . The air in

the cylinder is now compressed isothermally until the

radius of the bubble is halved. The pressure of air in the

cylinder now becomes 5 210 N/m .n× The surface tension

of soap film is –10.08Nm . Find the integer value of n.

11. A length of wire carries a steady current I. It is bent first

to form a circular plane coil of one turn. The same length

is now best more sharply to give double loop of smaller

radius. If the same current I is passed, the ratio of the

magnitude of magnetic field at the centre with its first

value is.

12. A current 1 amp is flowing in the sides of an equilateral

triangle of side 24 5 10 m.−⋅ × the magnetic field at the

centroid of the triangle in the unit of 5(10 T)− is.

13. There are two infinite long parallel straight current

carrying wires, A and B separated by a distance r (Fig.)

The current in each wire is I. The ratio of magnitude of

magnetic field at points P and Q when points P and Q lie

in the plane of wires is

14. A length of wire carries a steady current I. It is bent first

to form a circular plane coil of one turn. The same length

is now bent more sharply to give double loop of smaller

radius. If the same current I is passed, the ratio of the

magnitude of magnetic field at the centre with its first

value is



15. Work is defined as dot product of force and displacement

.W dW F dS= = ⋅∫ ∫

It is a scalar quantity. The total work

done will depend on the displacement and the force,

which may be constant or variable.

Thus in different situations of variable force applied in

column I, the final expression for work done can be

expressed as in column II.

Column I Column II

(A) Force constant in magnitude acts

at constant angle θ with direction

of motion

1. 21cos

2kx θ

(B) Force constant in magnitude acts

at angle θ which varies as θ kxθ = 2. sinF

Kθ θ

(C) Force varies with distance x as

F kx= but angle θ is constant

3. sinFx θ

(D) Force is constant in magnitude

but changes in direction with

changing angle and always acts

along radius of circular path

4. zero

a. A→1, B→2, C→3, D→4 b. A→3, B→2, C→1, D→4

c. A→4, B→3, C→2, D→1 d. A→2, B→1, C→4, D→3

16. Four charges 1 2 3, ,Q Q Q and 4Q of same magnitude are fixed

along the x-axis at 2 , ,x a a a= − − + and 2a+ respectively.

A positive charge q is placed on the positive y-axis at a

distance 0b > . Four options of the signs of these charges

are given in column I. the direction of the forces on the

charge q is given in column II. Match column I with

column II and select the correct answer using the code

given below the lists.

Column I Column II

(A) 1 2 3 4, , ,Q Q Q Q all positive 1. x+

(B) 1 2,Q Q positive;

3 4,Q Q negative 2. x−

(C) 1 4,Q Q positive; 2 3,Q Q negative 3. y+

(D) 1 3,Q Q positive;

2 4,Q Q negative 4. y−

q+

1

( 2 ,0)

Q

a−

(0, )b

2

( ,0)

Q

a−3

( ,0)

Q

a+4

( 2 ,0)

Q

a+

P

O

a

2a

r

r

P

I

Q

A B

I

Physics 26

a. A→3, B→1, C→4, D→2

b. A→4, B→2, C→3, D→1

c. A→3, B→1, C→2, D→4

d. A→4, B→2, C→1, D→3

17. The vibration of body can be under various types of

forces. The vibration are classified mentioned in column I

under the conditions mentioned in column II. Match the

type of vibrations in column I with conditions in

column II.

Column I Column II

(A) Free vibrations 1. A body vibrating in

viscous medium

(B) Forced vibrations 2. A body vibrating under

its natural restoring force

(C) Resonant vibrations 3. A body vibrating under

the influence of another

vibrating body

(D) Damped vibrations 4. A body vibrating with

its natural frequency

under the influence of

another vibrating body

of same frequency

a. A→1, B→2, C→3, D→4

b. A→4, B→3, C→2, D→1

c. A→2, B→3, C→4, D→1

d. A→3, B→4, C→1, D→2

18. To determine specific heat of different substances we use

different types of calorimeters. Can you match the type of

colorimeter used named in column I to the substance

whose specific heat is determined by corresponding

calorimeter in column II.

Column I Column II

(A) Regnault’s

calorimeter

1. To determine specific

heat of solids at very

low temperatures

(B) Joly’s differential

calorimeter

2. To determine specific

heat of solids or gases

at constant pressures

(C) Calender and

Barmer’s calorimeter

3. Used to measure specific

heat at constant volume

(D) Nernst’s vacuum

calorimeter

4. Specific heat of liquids

and gases at constant

pressure

a. A→1, B→2, C→3, D→4

b. A→2, B→3, C→4, D→1

c. A→3, B→4, C→1, D→2

d. A→4, B→3, C→2, D→1


27 Mock Test-2

ANSWER & SOLUTIONS

JEE-Main

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

a a d c c b c d d a

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

d a b c a d d b d c

21. 22. 23. 24. 25.

d d c c c

1. (a) 1 3 2

2 2[GM] [M L T M][V] [L T ]

[ ] [L]r

− −−= = = ⇒ [ ]

Z

kθ α =

Further [ ]p α= β

∴ [ ]Z

k

p p

α θβ = =

Dimensions of k θ are that to energy.

Hence, [ ]2 2

2[K ] [ML T ][MLT ]

[ ] [L]z

−−θ

α = = =

2. (a) Since, the block rises to the same heights in all the

four cases, from conservation of energy, speed of the block

at highest point will be same in all four cases. Say it is v0

Equation of motion will be

2

0mv

N mgR

+ =

or

2

0mvN mg

R= −

R (the radius of curvature) in first case is minimum.

Therefore, normal reaction N will be maximum in first

case.

Note: In the question it should be mentioned that all the

four tracks are frictionless. Otherwise, v0 will be different

in different tracks.

3. (d) dU

Fdx

= −

∴ . dU F dx= − or 3

0 ( ) ( )

x

U x kx ax dx= − − +∫

2 4

( )2 4

kx axU x = −

( ) 0U x = at x = 0 and

2kx

a=

( )U x = Negative for

2kx

a>

From the given function we can see that

F = 0 at x = 0 i.e., slope of U – x graph is zero at x = 0.

4. (c) If the density of cone be ρ, then its mass will be

( ) ( )2 3

1

1 162 4

3 3m R R Rπ ρ π ρ= =

And its centre of mass will be at a height 4

4 4

h RR= =

From O on the line of symmetry, i.e., 1y R=

Similarly the mass of the sphere

( )3 3

2 1

412 16 3

3m R R mπ ρ π ρ= = =

And its centre of mass will be at its centre 2 ,O i.e.,

2 4 5y R R R= + = (from O).

Now treating the sphere and cone as point masses with

their masses concentrated at their centres of mass

respectively and taking the line of symmetry as y-axis

with origin at O, the centre of mass of the toy is given by

1 1 2 2

1 2

CM

m y m yY

m m

+=

+1 1

1 1

3 54

3

m R m RR

m m

× + ×= =

+

i.e., centre of mass of the toy is at a distance 4R from O on

the line of symmetry, i.e., at the apex of the cone.

5. (c) The total energy of a particle in simple harmonic

motion is constant.

6. (b) Orbital speed of a satellite 0

GMv

r=

⇒ 4

2B A

A B

v r R

v r R= = =

⇒ ( )2 2 3 6B Av v v v= = × =

7. (c) Excess pressure inside soap bubble is inversely

proportional to the radius of bubble, i.e., ∆P ∝ 1/r′. This

means that bubbles A and C posses greater pressure inside

it than B. So the air will move from A and C, towards B.

2R

2R

C

O2 m2

y

O2

O1 O1 m1

O O

4R

v0

N + mg

Physics 28

8. (d) F Y A= ∆α∆θ

( ) ( )11 62.0 10 1 10−= × × × ( )( )51.1 10 40 20 44 N−× × − =

9. (d) 22 1 10

1

10logI

L LI

− =

⇒ 210

1

60 20 10logI

I− =

24 42 2

2

1 1

10 10I a

I a= ⇒ =

⇒ 21 2

1

100 : 1 : 100a

a aa

= ⇒ =

10. (a) dQ mc dt=

2

1

t

t

Q mc dt= ∫

( )2

1

40

2 3

20

100 2 10

t

t

t t dt

=−

=

= × + ×∫40

3 21

20

210

3 2

t t− = +

( ) ( ) ( ) ( )3 3 21 2 110 40 20 40 20

3 2

− = − + −

33.79 10= × cal = 3.79 kilocal

11. (d) A real gas behaves like an ideal gas at low pressure

and high temperature.

12. (a) When separation between the plates is doubled the

capacitance becomes one half i.e., 25C F′ = µF

Energy spent by battery qV=

( ) 2C V V C V′ ′= =

( )26 225 10 100 25 10− −= × × = × J

13. (b) mass/density

M MI

V= = ,

Given mass = 1gm = 310− kg

And density3

3 3 3

2 3 3

5 105 / 5 10 /

(10 )

kggm cm kg m

m

−

−

×= = = ×

Hence 7 3

3

6 10 5 103

10I

−

−

× × ×= =

14. (c) As the electric field is switched on, positive ion will

start to move along positive x-direction and negative ion

along negative x-direction. Current associated with motion

of both types of ions is along positive x-direction.

According to Fleming's left hand rule force on both types

of ions will be along negative y-direction.

15. (a) 3

2100 1 10 22 10

10

d nAdBdQ C

R R

φ −−× × ×

= = = = × C

16. (d) The current will lag behind the voltage when reactance

of inductance is more than the reactance of condenser.

Thus, 1

LC

ωω

> or LC

1>ω

or LC

nπ2

1> or rnn >

where, nr = resonant frequency.

17. (d) 2I a∝ ⇒ 1 1

2 2

100 10

1 1

a I

a I= = =

18. (b) The direction of propagation of electromagnetic wave

is given by the direction of ( ).E B×

Here, the em wave is

propagating along north. The electric field vector is acting

upwards, so the magnetic field vector will point towards

east.

19. (d) (Eion)Na2 2( ) (11) 13.6ion HZ E eV= = eV

20. (c) Fast neutrons can escape from the reaction. So as to

proceed the chain reaction. Slow neutrons are best.

21. (d) The component of force in vertical direction

1

cos cos60 5 2.5 N2

F F= θ = ° = × =

22. (d) There are 10 electrons and 10 protons in a neutral

water molecule.

So it's dipole moment is ( )2 10 (2 )p q l e l= =

Hence length of the dipole i.e,. distance between centres

of positive and negative charges is

20

12

19

6.4 102 4 10 m 4

10 10 1.6 10

pl pm

e

−−

−

×= = = × =

× ×4 pm

23. (c) For maximum power int 2extR R= = Ω

24. (c) 360

1 745

n°

= − =°

25. (c)34 15

0 6 10 1 6 10W hv −= = × × ⋅ × joule

34 15

19

6 10 1 6 10

1 6 10

−

−

× × ⋅ ×= =

⋅ ×eV = 6 eV

∴ 8 6 2kE hv W eV eV eV= − = − = 8 eV –6 eV = 2eV

y

x

60o

F sin 60o

F c

os

60

o

F

29 Mock Test-2


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

b c a,b,c a,c c,d c 6 3 2 7

11. 12. 13. 14. 15. 16. 17. 18.

6 6 3 4 b a a c

1. (b) tana

gθ =

tan 2dy

kxdx

θ = =

⇒ 2

ax

gk=

2. (c) sin cossmg mg=θ µ θ [when partied is just balanced]

⇒ tan su=θ ⇒

2

tan2

dy x

dx

θ = =

∴

2

0.52

x= ⇒ 1x =

⇒

1

6y =

3. (a, b, c) = +A A AE mgh K

⇒ =B BE K

⇒ = +C C CE mgh K

Using conservation of energy = =A B CE E E

⇒ >B CK K

⇒ >B AK K

⇒ ( ) ( ) 0− + − =A C A CMg h h K K ⇒−

− = C AA C

K Kh h

Mg

4. (a, c) 0 ( 45 )f = θ = °

So, for block to be at rest cos sinF mgθ = θ

LHS cos 1 cos45F θ = × °

1

2=

RHS 1 1

sin 0.1 102 2

mg θ = × × =

LHS = RHS)

If 45 ; sin cosmg Fθ > ° θ > θ

∴ Friction acts towards Q.

If 45θ < ° F cos sinmgθ > θ

∴ Friction acts towards P.

5. (c, d) Condition of translational equilibrium 1 2 2µ=N N

⇒ 2 1 1µ= =N N Mg , Solving 2

1 21 µ µ=

+mg

N

⇒ 2

1

1 21

µµ µ

=+

mgN Applying torque equation about corner (left)

point on the floor

⇒ 1 1cos sin cos

2mg N N= +ℓ

ℓ ℓ1θ θ µ θ

Solving 1 2

2

1tan

2

−µ µθ =

µ

6. (c) Velocity will exchange after each collision

7. (6) 2 22

43

ρ π= ∫I r r dr

⇒ 2 2( )( )( )∝ ∫AI r r r dr

⇒ 5 2 2( )( )( )∝ ∫BI r r r dr

∴ 6

10=B

A

I

I

8. (3)' 6 ' 2

;11 3

g

g

ρρ

= =

Hence, ' 3 6

22

R

R=

⇒ 2

esc

2

sec

' ' ' 3

V 11

V R

R

ρρ

∝ = ⇒ esc' 3km/s.v =

9. (2) At height R from the surface of planet acceleration due

to planet’s gravity is 1

4th in comparison to the value at the

surface

So, 2 21 1and 0

2 2esc

GMm GMm GMnmv mv

R R R R− + = − − + =

+

∴ 2escv v=

2 /3πV

st1 collision

A 2V V

V 2V

2V

nd2 collision

2 /3π

2 /3π

F

θmg

F

Fcosθ

MgsinθMg cosθ

fs N

y

N cosθ

ma

N

θ N

θ

sin θ

x

ma

θ θ

Physics 30

10. (1)

1( 10)d dxℓ = θ −α d∆ = ∫ℓ ℓ

∵ 10 130

1x

θ − = ⇒ 10 130xθ = +

1

1

0

(130 )x dxα∆ = ∫ℓ 2

1130

2

xα∆ =ℓ

5 1

130 1.2 102

−∆ = × × ×ℓ = 0.78 mm 1≃

11. (6) 2

0

416N / m

A

A

TP P

R= + =

⇒ 2

0

412N / m

B

B

TP P

R= + =

⇒

3

6

= =

B B B

A A A

n P R

n P R

12. (6) 3 2

0 98 10 N / mP P ghρ= − = ×

⇒ 0 0 =PV PV

⇒ 5 310 [ (500 )] 98 10 [ (500 200)]− = × −A H A

⇒ 206 mmH =

Level fall = 206 – 200 = 6 mm

13. (3) 06 L grv V Vgπη ρ ρ+ =

6( )

6 ( )

Q QP P P L P

Q P P Q Q L Q

rV V V g

V r V V

πηρ ρπη ρ ρ−

= ×−

3

3

.(8 0.8)

. (8 1.6)

Q QP

P P Q

rr

r r

η

η−

= ×−

2

7.2

6.4

QP

Q P

r

r

η

η

= × ×

7.2 24 3

6.4 3= × × =

14. (4) YA

mLω =

15. (b) Using work energy theorem

21sin 30

2fmg R W mv° + =

⇒ 2

200 1502

v− =

⇒ 10 m/sv =

16. (a)2

cos60mv

N mgR

− ° =

⇒ 5

5 7.5 N2

N = + =

17. (a) 1

1 2 4f f s−− =

18. (c) 1 2 200 m / sv v= = m/s


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

d a a,c a d c 8 7 3 8

11. 12. 13. 14. 15. 16. 17. 18.

4 4 8 4 b a c a

1. (d) Let m mass of fat is used.

7 1

(3.8 10 ) 10(9.8)(1)(1000)5

m × =

3

9.8 5

3.8 10m

×=

× 312.89 10−= × kg

2. (a) Linear momentum remains constant if net external

force on the system of particles is zero.

3. (a, c) By conservation of linear momentum

5 2u v= − . . .(i)

By Newton’s experimental law of collision

2u v= + . . .(ii)

Using (i) and (ii) we have 1m/sv = and 3 m/su =

Kinetic energy of the centre of mass

2

system cm

10.75J

2m v= =

4. (a) 2 22 3

5 2=MR Mr ⇒

2

15=

Rr

5. (d) 2

max maxm r Tω =

⇒ max

max

324

0.5 0.5

T

mrω = =

×1296 36 rad/s= =

5 kg

before collision

1 kg

u

after collision

1 kgV

5 kg

H 500 mm

500 mm

200 mm

RBR

4cm

BA

AR

2cm

28 N / m

P 1m ,Q R 1m S

10 C° 140 C°400 C°

2,k α12 ,k α

31 Mock Test-2

6. (c) 2 ˆ( )OL m r kω=

Direction of PL

varies with time as its direction will be

perpendicular to string i.e. changing with time.

OL =

Constant , PL

= Variable

7. (8) Here, 1 500mm 0.5mλ = =

2 1 20.25 m; 16 m/s, ?λ υ υ= = =

As frequency of body is fixed, say n, therefore,

1 1 2 2;n nυ λ υ λ= =

⇒ 2 2

1 1

0.25 1

0.5 2

υ λυ λ

= = =

⇒ 12

2

168m/s

2

υυ

υ= = =

8. (7) Here, . 2, 40cm, ?m h n= = =

Time taken by the plate to fall down,

2 2 40 2

sec980 7

ht

g

×= = =

Frequency of fork, 2

7Hz.2 / 7

mn

t= = =

9. (3) The pressur at the bottom,0P P h gρ= +

Where, 0p = atmospheric pressure, therefore,

0

1h g

PP

ρ= + atmosphere

20

1 310

g

g

ρρ

= + = atmosphere.

10. (8) Here, 5 2 410 N/m , 2.4 10 m; 0.08N/mp R S

−= = × =

Initial pressure inside the soap bubble in air cylinder,

1

4.S

P PR

= +

Let 1V be the initial volume of the soap bubble;

3

1

4

3V Rπ=

After compression, volume of the soap bubble;

3

12

4

3 2 8

R VV π = =

If is the air pressure after compression in the cylinder,

then pressure inside the bubble is

2

4 8

( / 2)

S SP P P

R R

′ ′= + = +

Using Boyle’s law, we have, 1 1 2 2P V PV=

or 11

4 8

8

S S VP V P

R R

′+ = +

or 4 1 8

9

S SP P

R R

′+ = +

or 5

4

24 24 0.088 8 10

2.4 10

SP P

R −

+′ = + = × +×

5 58.08 10 10n= × = ×

∴ 0.08 8.n = ≈

11. (4) When wires is taken in the form of one turn circular

coil, then length, 2l rπ= or , 12

lr n

π= = Magnetic field

induction at the centre of circular coil due to current I is

0 0 02 2 1

4 4 ( / 2 )

InI IB

r l l

µ µ µ ππ ππ π π

× ×= = =

When wire is taken in the form of double loop, then

12 2l rπ= ×

or 1

4

Ir

π= and 2n =

∴ Magnetic field induction at the centre of the circular coil.

01

1

2

4

n IB

r

µ ππ

× ×= 0

0

2 24

4 ( / 4 )

I I

l l

µ π πµ

π π× ×

= = ×

∴ 1 4B

B=

12. (4) Refer Fig.

The magnetic field induction at the centriod O due to

current I through one side BC of the triangle will be

0

1 1 2(sin sin )

4

IB

r

µ= θ + θ

π

It will be acting perpendicular to the plane of triangle

upwards. Total magnetic field induction at O due to

current through all the three sides of the triangle will be

0

1 1 2

33 [sin sin ]

4

IB B

µ= = θ + θ

π π

Here, 1 2

1 , 60I A= θ = ° = θ and

/ 2

tan 60 3

BD ar OD= = =

°

A

C

I

D B I

O

I

a

60°r

60°

OL

PLθ

Physics32

2

2

4 5 10.

3 2 3

am

−⋅ ×= =

∴ 7

2

13 10

(4 5 10 / 2 3)B −

−= × ×

⋅ ×[sin 60 sin 60 ]× ° + °

On solving, 54 10 T.B −= ×

13. (8) Magnetic field at P due to currents in two wires will be

acting perpendicular to the plane of wires, upwards and is

given by.

Magnetic field at Q due to current in A is perpendicular to

the plane of wire upwards and due to current in B is

perpendicular to the plane of wire downwards and is given

by

8.

14. (4) When wire is taken in the form of one turn circular

coil, then length

or

Magnetic field induction at the centre of circular coil due

to current I is

When wire is taken in the form of double loop, then

or and n = 2

Magnetic field induction at the centre of the circular coil,

⇒ = 4.

15. (b) A→3, B→2, C→1, D→4

Displacements while angle θ between and is

constant

θ henced

Kdx=

θ or

ddx

k=

θ

cos sin

d F

K K=

θθ θ

21cos cos

2Kxθ = θ

at every point

Qθ = 90°

16. (a) A→3; B→1; C→4; D→2

17. (c) A→2, B→3, C→4, D→1

When a body is disturbed from its mean position and left

to vibrate under restoring force, e.g., A tunning force

struck with rubber pad, vibrations are called free or

natural vibrations. When a body is mode to vibrate with

another vibrating body placed nearby, e.g., stem of a

vibrating tunning fork when pressed on top of sonometer

makes the sonometer wire to vibrate having forced

vibrations. When a body is forced to vibrate by another

vibrating body the vibrations are forced on other body.

But if frequency of forced vibrations is equal to the

natural frequency of forced body then vibrations of forced

body are called resonant vibrations. When a body is

vibrating under the action of viscous force, gradually the

amplitude of vibration decreases because energy of

vibrating body is dissipated as work done against viscous

force in the form of heat etc. and body gradually stops

vibrating. Such vibratious are called damped vibrations.

18. (a) A→1, B→2, C→3, D→4

The particular names of calorimeters are after the names

of scientists who designed them for specific heat

measurements in different conditions.

( ) ( )0 0 022 2

4 / 2 4 / 2P

II IB

r r r

µ µ µπ π π

= + =

0 0 02 2

4 2 4 4Q

I I IB

r r r

µ µ µπ π π

= + =

∴( )( )

0

0

2 /

/ 4

P

Q

I rB

B I r

µ π

µ π= =

2l rπ=

, 12

lr n

π= =

( )0 0 02 2 1

4 4 / 2

InI IB

r l l

µ µ µ ππ ππ π π

× ×= = =

12 2l rπ= ×1

4

lr

π=

∴

( )0 0

1 0

1

2 2 24

4 4 / 4

n I I IB

r l l

µ µπ π πµ

π π π× × × ×

= = = ×

1B

B

cosW F S Fx θ= ⋅ =rr

∴ S x=r r

Fr

xr

,kxθ =

∴ dW F dS= ⋅∫ ∫uurr

F dx F= ⋅ = ⋅ ⋅ =∫ ∫uurr

W dW F dx= = ⋅∫ ∫uurr

Kx dx= ⋅∫uur uur

K xdx Kx= =∫F x⊥r r

∴ 0dW F dx= ⋅ =uurr

cos90 0Fdx= °=

33Mock Test-3



1. If ˆˆ ˆ2 4 5A i j k= + −ur

the direction of cosines of the vector

Aur

are:

a. 45

5and

45

4,

45

2 − b.

45

3and

45

2,

45

1

c. 45

4and0,

45

4 d.

45

5and

45

2,

45

3

2. A particle is projected vertically upwards and it is at a

height h after 2 seconds and again after 10 seconds. The

height h is:

a. 196 m b. 98 m

c. 9.8 m d. 19.8 m

3. An insect crawls up a hemispherical

surface very slowly (see the figure).

The coefficient of friction between

the surface and the insect is 1/3. If

the line joining the centre of the hemispherical surface to

the insect makes an angle α is given

a. cot 3α = b. tan 3α =

c. sec 3α = d. cosec 3α =

4. An ideal spring with spring constant k is hung from the

ceiling and a block of mass M is attached to its lower end.

The mass is released with the spring initially unscratched.

Then the maximum extension in the spring is

a. b.

c. d.

5. A particle executes simple harmonic motion between

and The time taken for it to go from

0 to A/2 is and to go from A/2 to A is Then:

a. b.

c. d.

6. A large number of liquid drops each of radius r coalesce

to form a single drop of radius R. The energy released in

the process is converted into the kinetic energy of the big

drop so formed. The speed of the big is (Given surface

tension of liquid is T, density of liquid is ρ)

a. b.

c. d.

7. A solid sphere of radius R made of material of bulk

modulus K is surrounded by a liquid in a cylindrical

container. A massless piston of area A floats on the

surface of the liquid when a mass m is placed on the

piston to compress the liquid, the fractional change in the

radius of the sphere, is:

a. b.

c. d.

8. The wavelength of light observed on the earth, from a

moving star is found to decreases by 0.05%. Relative to

the earth, the star is:

a. moving away with a velocity of 5

1.5 10× m/s

b. coming closer with a velocity of 51.5 10× m/s

c. moving away with a velocity of41.5 10× m/s

d. coming closer with a velocity of 4

1.5 10× m/s

9. An ideal gas expands isothermally from a volume to

and then compressed to original volume

adiabatically. Initial pressure is and final pressure is

The total work done is W. Then,

a. b.

c. d.

10. At NTP one mole of diatomic gas is compressed

adiabatically to half of its volume ( ).=γ 1.40 The work

done on the gas will be

a. 1280 J b. 1610 J

c. 1792 J d. 2025 J

11. Two identical conducting rods are first connected

independently to two vessels, one containing water at

100°C and the other containing ice at 0ºC. In the second

case, the rods are joined end to end connected to the same

vessels. Let and gram per second be the rate of

melting of ice in the two cases respectively. The ratio

is

a. b.

c. d.

4Mg

k

2Mg

k

Mg

k 2

Mg

k

x A= − .x A= +

1T 2.T

1 2T T< 1 2T T>

1 2T T= 1 22T T=

1 1T

r Rρ −

2 1 1T

r Rρ −

4 1 1T

r Rρ −

6 1 1T

r Rρ −

R

R

δ

mg

AK 3

mg

AK

mg

A 3

mg

AK

1V

2V 1V

1p

3.p

3 1, 0> >p p W 3 1, 0< <p p W

3 1, 0p p W> < 3 1, 0= =p p W

( )0.42 1.40 :=

1q 2q

1 2/q q

1

2

2

1

4

1

1

4

α

Physics34

12. The energy density (energy per unit volume) in an electric

field caused by a point charge falls off with the distance

from the point charge as:

a. b.

c. d.

13. The needle of a deflection galvanometer shows a

deflection of 60° due to a short bar magnet at a certain

distance in position. If the distance is doubled, the

deflection is

a. b.

c. d.

14 If power factor is in a series RL circuit .

ac mains is used then L is

a. 3

henryπ

b. henryπ

c. henry2

π d. None of these

15. The amplitude ratio of two superposing waves is 2 : 1.

The ratio of maximum and minimum intensities is:

a. 1 : 1 b. 2 : 1

c. 4 : 1 d. 9 : 1

16. The electric and the magnetic field, associated with an

e.m. wave, propagating along the + z-axis, can be

represented by

a. b.

c. d.

17. If the wavelength of the first line of the Balmer series of

hydrogen is 6561 Å, the wavelength of the second line of

the series should be

a. 13122 Å b. 3280 Å c. 4860 Å d. 2187 Å

18. In the nuclear reaction X,++

β what does

X stand for

a. An electron b. A proton

c. A neutron d. A neutrino

19. The symbol given in figure represents

a. NPN transistor

b. PNP transistor

c. Forward biased PN junction diode

d. Reverse biased NP junction diode


20. For television broadcasting, the frequency employed is

normally

a. 30–300 MHz b. 30–300 GHz

c. 30–300 kHz d. 30–300 Hz

21. The potential difference V and current i flowing through

an appliance in an ac circuit are given by

and power dissipated in

the appliance is

a. 0 W b. 10 W

c. 5 W d. 2.5 W

22. A wire has a mass (03. ± 0.003) g, radius (0.5 ± 0.005) mm

and length (6 ± 0.06) cm. The maximum percentage error

in the measurement of its density is:

a. 1 b. 2

c. 3 d. 4

23. An aircraft with a wing-span of 40 m flies with a speed of

1080 km h–1 in the eastward direction at a constant

altitude in the northern hemisphere, where the vertical

component of earth's magnetic field is 48.3 10 T.−× Then

the e.m.f. that develops between the tips of the wings is:

a. 0.5 V b. 0.35 V

c. 1 V d. 2.1 V

24. If an observer is walking away from the plane mirror with

3 m/sec. Then the velocity of the image with respect to

observer will be

a. b.

c. d.

25. The eccentricity of earth’s orbit is 0.0167. The ratio of its

maximum speed in its orbit to its minimum speed is:

a. 2.507 b. 1.033

c. 8.324 d. 1.000

1/ r 21/ r

31/ r41/ r

Atan

1 3sin

8−

1 3cos

8−

1 3tan

8−

1 3cot

8−

1

2100R = Ω

0 0ˆ ˆ,E E i B B j = =

r r

0 0ˆ ˆ,E E k B B i = =

r r

0 0ˆ ˆ,E E j B B i = =

r r

0 0ˆˆ,E E j B B k = =

r r

BC +→ 115

116

5cos volt, 5sin ampV t i tω ω= =

6 m/sec 6 m/sec−

12 m/sec 3 m/sec

E C

B

Space for Rough Work

35 Mock Test-3

JEE ADVANCE PAPER-I



1. In a large building, there are 15 bulbs of 40 W, 5 bulbs of

100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage

of the electric mains is 220 V. The minimum capacity of

the main fuse of the building will be:

a. 12 A b. 14 A

c. 8 A d. 10 A

2. When 5V potential difference is applied across a wire of

length 0.1 m, the drift speed of electrons is 4 12.5 10 ms .− −× If the electron density in the wire is

28 38 10 m ,−× the resistivity of the material is close to:

a. 81.6 10 m−× Ω b. 71.6 10 m−× Ω

c. 61.6 10 m−× Ω d. 51.6 10 m−× Ω 3. A galvanometer having a coil resistance of 100 Ω gives a

full scale deflection, when a current of 1 mA is passed

through it. The value of the resistance, which can convert

this galvanometer into ammeter giving a full scale

deflection for a current of 10 A, is:

a. 0.01 Ω b. 2 Ω

c. 0.1 Ω d. 3 Ω 4. The temperature dependence of resistances of Cu and

undoped Si in the temperature range 300–400 K, is best

described by:

a. Linear increase for Cu, linear increase for Si

b. Linear increase for Cu, exponential increase for Si

c. Linear increase for Cu, exponential decrease for Si

d. Linear decrease for Cu, linear decrease for Si

5. Two ideal batteries of emf 1V and 2V and three resistances

1 2,R R and are 3R connected as shown in the figure. The

current in resistance 2R would be

zero if

a. 1 2=V V and 1 2 3= =R R R

b. 1 2=V V and 1 2 32= =R R R

c. 1 22=V V and 1 2 32 2= =R R R d. 1 22 =V V and 1 2 32 = =R R R

6. Heater of an electric kettle is made of a wire of length L

and diameter d. It takes 4 minutes to raise the temperature

of 0.5 kg water by 40 K. This heater is replaced by a new

heater having two wires of the same material, each of

length L and diameter 2d. The way these wires are

connected is given in the options. How much time in

minutes will it take to raise the temperature of the same

amount of water by 40 K?

a. 4 if wires are in parallel

b. 2 if wires are in series

c. 1 if wires are in series

d. 0.5 if wires are in parallel



7. A current 1 amp is flowing in the sides of an equilateral

triangle of side 24.5 10 m−× . The magnetic field at the

centroid of the triangle in the units of ( )510 T− is

8. If the maximum values of signal and carrier waves are 4 V

and 5V respectively, the percentage of amplitude

modulation is a × 10%. What is the value of a ?

9. A signal wave of frequency 4.5 kHz is modulated with a

carrier wave of frequency 3.45 MHz. The bandwidth of

FM wave is kHz is

10. What is the maximum usuable frequency (in MHz) for

E-layer of atmosphere having critical frequency 4 MHz,

when the angle of incidence is 60°?

11. A T.V. Tower has a height 100 m. In order to triple its

coverage range, the height of tower to be increased is 210 m.a × What is the integer value of a?

12. A microwave telephone link operating at the central

frequency of 10 GHz has been established. If 2% of this is

available for microwave communication channel and each

telephone is allotted a bandwidth of 8 kHz, the number of

telephone channels which can be simultaneously granted

is 2.5 10 .a× What is the integer value of a ?

13. A uniform circular disc of mass 1.5 kg and radius is

initially at rest on a horizontal frictionless surface. Three forces of equal magnitude 0.5F N= are applied

simultaneously along the three sides of an equilateral triangle XYZ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the

angular speed of the disc in –1rad s is

F

F

FX

Z Y

O

2V

1V1R

2R

3R

Physics 36

14. A horizontal circular platform of radius 0.5 m and mass

0.45 kg is free to rotate about its axis. Two mass less

spring toyguns, each carrying a steel ball of mass 0.05 kg

are attached to the platform at a distance 0.25 m from the

centre on its either sides along its diameter (see figure).

Each gun simultaneously fires the balls horizontally and

perpendicular to the diameter in opposite directions. After

leaving the platform, the balls have

horizontal speed of 19 ms− with

respect to the ground. The

rotational speed of the platform in –1rad s after the balls leave the platform is





In a thin rectangular metallic strip a constant current I flows

along the positive x-direction, as shown in the figure. The

length, width and thickness of the strip are l, w and d,

respectively. A uniform magnetic field B

is applied on the

strip along the positive y-direction. Due to this, the charge

carriers experience a net deflection along the z-direction. This

results in accumulation of charge carriers on the surface PQRS

and appearance of equal and opposite charges on the face

opposite to PQRS. A potential difference along the z-direction

is thus developed. Charge accumulation continues until the

magnetic force is balanced by the electric force. The current is

assumed to be uniformly distributed on the cross-section of the

strip and carried by electrons.

15. Consider two different metallic strips (1 and 2) of the

same material. Their lengths are the same, widths are w1

and w2 and thicknesses are d1 and d2, respectively. Two

points K and M are symmetrically located on the opposite

faces parallel to the x-y plane (see figure). V1 and V2 are

the potential differences between K and M in strips 1 and

2, respectively. Then, for a given current I flowing

through them in a given magnetic field strength B, the

correct statement(s) is(are)

a. If w1 = w2 and d1 = 2d2, then V2 = 2V1

b. If w1 = w2 and d1 = 2d2, then V2 = V1

c. If w1 = 2w2 and d1 = d2, then V2 = 2V1

d. If 1 22w w= and 1 2 ,d d= then 2 1V V=

16. Consider two different metallic strips (1 and 2) of same

dimensions (lengths ,ℓ width w and thickness d) with

carrier densities n1 and n2, respectively. Strip 1 is placed

in magnetic field B1 and strip 2 is placed in magnetic field

B2, both along positive y-directions. Then V1 and V2 are

the potential differences developed between K and M in

strips 1 and 2, respectively. Assuming that the current I is

the same for both the strips, the correct option(s) is(are)

a. If 1 2B B= and 1 22 ,n n= then 2 12V V=

b. If 1 2B B= and 1 22 ,n n= then 2 1V V=

c. If 1 22B B= and 1 2 ,n n= then 2 10.5V V=

d. If 1 22B B= and 1 2 ,n n= then 2 1V V=


The capacitor of capacitance C can be charged (with the help of

a resistance R) by a voltage

source V, by closing switch 1S

while keeping switch 2S open.

The capacitor can be

connected in series with an

inductor ‘L’ by closing switch

2S and opening 1S

17. Initially, the capacitor was uncharged. Now, switch S1 is

closed and S2 is kept open. If time constant of this circuit

is τ, then

a. after time interval τ, charge on the capacitor is CV/2

b. after time interval 2 ,τ charge on the capacitor is 2(1 )CV e−−

c. the work done by the voltage source will be half of the

heat dissipated when the capacitor is fully charged

d. after time interval 2 ,τ charge on the capacitor is 1(1 )CV e−−

18. After the capacitor gets fully charged, S1 is opened and S2 is closed so that the inductor is connected in series with

the capacitor. Then

a. at 0,t = energy stored in the circuit is purely in the

form of magnetic energy

b. at any time 0,t > current in the circuit is in the same

direction

c. at 0,t > there is no exchange of energy between the

inductor and capacitor

d. at any time 0,t > instantaneous current in the circuit

may C

VL

y

x

z

I

Q P

S

I W

K•

M•R

d

V

C

R S1

S2 L

l

37 Mock Test-3




1. The figure shows certain wire segments joined together to

form a coplanar loop. The loop is placed in a

perpendicular magnetic field in the direction going into

the plane of the figure. The magnitude of the field

increases with time. 1I and 2I are the currents in the

segments ab and .cd Then,

a. 1 2>I I

b. 1 2<I I

c. 1I is in the direction ba and 2I is in the direction cd

d. 1I is in the direction ab and 2I is in the direction dc

2. A thin flexible wire of length L is connected to two

adjacent fixed points and carries a current I in the

clockwise direction, as shown in the figure. When the

system is put in a uniform magnetic field of strength B

going into the plane of the paper, the wire takes the shape

of a circle. The tension in the wire is

a. IBL b. IBL

π c.

2

IBL

π d.

4

IBL

π

3. Along insulated copper wire is closely wound as a spiral

of 'N' turns. The spiral has inner radius 'a' and outer radius

'b'. The spiral lies in the X-Y plane and a steady current 'I'

flows through the wire. The Z component of the magnetic

field at the center of the spiral is

a. 0 bIn

2( )

NI

b a a

−

µ b. 0 +

In2( ) –

NI b a

b a b a

−

µ

c. 0 In2

NI b

b a

µ d. 0 In

2

NI b+a

b b – a

µ

4. A loop carrying current I lies in the x-y plane as shown

in the figure. The unit vector k is coming out of the

plane of the paper. The magnetic moment of the current

loop is

a. 2 ˆa Ik b. 2 ˆ12

π +

a Ik

c. 2 ˆ12

π − +

a Ik d. 2 ˆ(2 1)π + a Ik

5. An infinitely long hollow conducting cylinder with inner

radius R/2 and outer radius R carries a uniform current

density along its length. The magnitude of the magnetic

field, | |B as a function of the radial distance r from the

axis is best represented by.

a. b.

c. d.

6. A circular loop of radius 0.3 cm lies parallel to a much

bigger circular loop of radius cm.20 The centre of the

small loop is on the axis of the bigger loop. The distance

between their centres is .cm15 If a current of 2.0 A flows

through the smaller loop, then the flux linked with bigger

loop is

a. weber101.9 11−× b. weber106 11−×

c. weber103.3 11−× d. weber106.6 9−×

r

B

R/2 R r

B

R/2 R

r

B

R/2 R r

B

R/2 R

y

a x

a

I

X

Y

I a b

c d

a b

Physics 38



7. The isotope 125 B having a mass 12.014 u undergoes

β - decay to 12 126 6C. C has an excited state of the nucleus

12 *6( C ) at 4.041 MeV above its grounds state. If

125 B decays to 12 *

6 C , the maximum kinetic energy of the

β - particle in units of MeV is:

(1 931.5MeV / ,u c= c2, where c is the speed of light in

vacuum)

8. A binary star consists of two stars A (mass 2.2Ms) and B

(mass 11Ms), where Ms is the mass of the sun. They are

separated by distance d and are rotating about their centre

of mass, which is stationary. The ratio of the total angular

momentum of the binary star to the angular momentum of

star B about the centre of mass is

9. Two homogeneous spheres A and B of masses m and 2m

having radii 2a and a respectively are placed in contact.

The ratio of distance of c.m. from first sphere to the

distance of c.m. from second sphere is:

10. A non-uniform thin rod of length L is placed along X-axis

so that one of its ends is at the origin. The linear mass

density of rod is 0xλ λ= ⋅ The centre of mass of rod

divides the length of the rod in the ratio:

11. A sphere of mass 5 kg and diameter 2 m rotates about a

tangent. What is its moment of inertia?

12. A uniform rod of length 1 m and mass 0.5 kg rotates at

angular speed of 6 rad/sec about one of its ends. What is

the KE of the rod?

13. A particle performing uniform circular motion has angular

momentum L. When its angular velocity is doubled and

KE is also doubled, the new angular momentum becomes

x times. What is x?

14. The fundamental frequency of a closed organ pipe is equal

to first overtone frequency of an open organ pipe. If the

length of the open pipe is 36 cm, what is the length

(in cm) of the closed pipe?



15. Match Column I with Column II and select the correct

answer using the codes given below the Columns:

Column I Column II

(A) Boltzmann Constant 1. 2 1[ML T ]−

(B) Coefficient of viscosity 2. 1 1[ML T ]− −

(C) Plank Constant 3. 3 1[MLT K ]− −

(D) Thermal conductivity 4. 2 2 1[ML T K ]− −

a. A-4; B-1, 4; C- 4; D-2

b. A-1; B- 1,4; C-4; D-4

c. A-4; B-1,4; C-4; D-2

d. A-1; B- 1,4; C-2; D-4

16. Column I give a list of possible set of parameters

measured in some experiments. The variations of the

parameters in the form of graphs are shown in Column II.

Match the set of parameters given in Column I with the

graph given in Column II.

Column I Column II

(A) Potential energy of a

simple pendulum (y axis)

as a function of

displacement (x axis)

1.

(B) Displacement(y axis) as a

function of time (x axis)

for a one dimensional

motion at zero or constant

acceleration when the

body is moving along the

positive x-direction

2.

(C) Range of a projectile

(y-axis) as a function of its

velocity (x-axis) when

projected at a fixed angle

3.

(D) The square of the time

period (y-axis) of a simple

pendulum as a function of

its length (x -axis)

5.

a. A-4; B- 1,4; C- 4; D- 2

b. A-1; B- 1,4; C- 4; D- 4

c. A-2; B- 1,4; C- 4; D- 2

d. A-1; B- 1,4; C- 2; D- 4

17. Motion is defined as rate of change of position. In the

motions described in column I, match the facts about that

motion in column II for the changes in position vector.

y

x O

y

x O

y

x O

y

x O

39 Mock Test-3

Column I Column II

(A) Magnitude only 1. A coin dropped from

roof of house

(B) Direction only 2. A coin thrown at any

angle with horizontal

(C) Both in magnitude and

direction

3. A coin held in your

hand

(D) Remains invariant 4. A coin in rotated in

circular path with

variable speed

a. A-1; B-2; C-3; D-4

b. A-4; B-2; C-3; D-1

c. A-1; B-4; C-2; D-3

d. A-3; B-2; C-4; D-1

20. Match the statement of Column with those in Column II:

Column I Column II

(A) In any Bohr orbit of the

hydrogen atom, the ratio 1.

1

2−

of kinetic energy to

potential energy of the

electron is

(B) The ratio of the kinetic

energy to the total energy

of an electron in a Bohr

orbit is

2. 2

(C) In the lowest energy level

of hydrogen atom, the

electron has the angular

momentum

3. 2

h

π

(D) Ratio of the wavelengths

of first line of Lyman

series and first line of

Balmer series is

4. 5 : 27

a. A-1; B-3,4; C-4; D-2,3

b. A-1; B-3,4; C-4; D-3

c. A-2; B-3,4; C-1; D-3

d. A-1; B-2,4; C-3; D-4


Physics 40

ANSWER & SOLUTIONS

JEE-Main

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

a b a b b d b b c c

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

c d c a d a c d a a

21. 22. 23. 24. 25.

a d c a b

1. (a) kjiA ˆ5ˆ4ˆ2 −+=

∴ 45)5()4()2(|| 222 =−++=A

∴ 2

cos45

α = , 4

cos45

β = , 5

cos45

−γ =

2. (b) 21

2h ut gt= −

⇒ ( )212 10

2h u g= × − × ( )21

10 102

u g= × − ×

⇒ 2 2 10 50u g u g− = − ⇒ 6u g=

∴ 2 2 2 6 2 10h u g g g g= − = × − = = 98 m

3. (a) Equilibrium of insect gives

cos N mg= α . . . (i)

sin N mg= αµ . . . (ii)

From equation (i) and (ii) we get 1

cot 3α = =µ

4. (b) Let x be the maximum

extension of the spring. From

conservation of mechanical energy

decrease in gravitational potential

energy

= increase in elastic potential energy

∴ 21

2Mg x kx=

or

2Mgx

k=

5. (b) For block P, friction will provide the necessary

restoring force.

∴ 2

maxf m Aω= with 2

2

k k

m m mω = =

+

Hence, max2 2

k kAf m A

m

= =

6. (d) Work done, W T A= ∆

or 2 2 2 2[(4 ) 4 ] .4 [ ]nW T R R T nr Rπ π π= − = −

Where 3 3 24 4 1and

3 3 2n r R W mπ π υ× = =

∴ 2 2 21.4 ( )

2m T nr Rυ π= −

or 3 2 2 21 4.4 ( )

2 3R T nr Rπ ρυ π× = −

or 2 2

3 3

6T nr R

R Rυ

ρ

= −

2

3

6 1 6 1 1T nr T

nr R r Rρ ρ = − = −

7. (b) /

/ /

P mg A

V V V V

∆=

∆ ∆V mg

V AK

∆= . . .(i)

But for a sphere 4

3V = π 2R

Differentiating 4

( )3

V =δ π δ23R R

⇒

4)

334

3

V=

V R=

π(3 δδ δ

π

2

3

R RR

R

∴ From (i) 3

V mg R mg=

R AK R AK⇒ =

3δ δ

8. (b) 0.05

100λ λ∆ = , As

v

cλ λ∆ =

⇒ 0.05

100

v

cλ λ= or 45 10v c−= ×

or 4 8 55 10 3 10 1.5 10v −= × × × = × m/s

As λ decreases, the star is approaching the observer.

9. (c) Slope of adiabatic process

at a given state (p, V, T) is

more than the slope of

isothermal process. The

corresponding p-V graph for

the two processes is as shown

in figure. In the figure, AB is isothermal and BC is

adiabatic.

ABW = positive (as volume is increasing) and

BCW = negative (as volume is decreasing) plus,

,BC ABW W> as area under p-V graph gives the work done.

Hence, 0AB BCW W W+ = <

From the graph itself, it is clear that 3 1.p p>

Hence, the correct option is

Note: At point B, slope of adiabatic (process BC) is

greater than the slope of isothermal (process AB).

N

cosmg αmg

sinmg α

αNµ

α

x

k

v = 0

v = 0

M

M

A

V

p

B

C

41 Mock Test-3

10. (c) 11 1 2 2 2 1

2

VP V P V P P

V

γ

γ γ = ⇒ =

work ( )2 2 1 11

RP V P V

γ= −

−

1

11

2

11

1

VP

V

γ

γ

− = −

−

( )0.405 3110 22.4 10 2 1

1 1.40− = × × × × −

−

[ ]2122.4 10 1.32 1 1792

0.40= × × − = J

11. (c)dQ dm

Ldt dt

=

or Temperature difference

Thermal resistance

dmL

dt

=

or 1

Thermal resistance

dm

dt∝ ⇒

1q

R∝

In the first case rods are in parallel and thermal resistance

is 2

R while in second case rods are in series and thermal

resistance is 2R.

⇒ 1

2

2 4

/ 2 1

q R

q R= =

12. (d) Energy u ∝ E2 and 2

1u

r∝ for a point charge; so

4

1u

r∝

13. (c) For short bar magnet in tan A-position

03

2tan

4

MH

d

µπ

= θ

. . . (i)

When distance is doubled, then new deflection θ ′ is given by

03

2tan

4 (2 )

MH

d

µπ

= ′θ

. . . (ii)

∴ tan

tan

′θθ

tan 1

tan 8θ′

=

⇒ –1tan tan 60 3 3tan tan

8 8 8 8

° ′ ′ = = = ⇒

θθ θ =

14. (a) 1

cos 602⇒ = °φ = φ

3

tan 60 HL

LR

° = ⇒ =ω

π

15. (d) ( )( )

2 2

1 2max2

min 1 2

2 1 9

2 1 1

a aI

I a a

+ + = = = − −

16. (a) In the given question, the electromagnetic wave is

propagating along z+ axis. In e.m. wave, the electric field

( )E

and magnetic field ( )B

are perpendicular to each other

and also perpendicular to the direction of propagation of

wave,

So, 0ˆE E i=

and 0

ˆ.B B j=

17. (c) The wavelength of spectral line in Balmer series is

given by 2 2

1 1 1

2R

nλ

= −

For first line of Balmer series, n = 3

⇒ 2 2

1

1 1 1 5

2 3 36

RR

λ = − =

; For second line n = 4.

⇒ 2 2

2

1 1 1 3

2 4 16

RR

λ = − =

∴ 21

1

20 206561 4860

27 27Å

λλ

λ= ⇒ = × = Å

18. (d) 11 116 5C B→ + β+ + γ because β+ 0

1 e=

19. (a) The base is always thin

20. (a) VHF (Very High Frequency) band having frequency

range 30 MHz to 300 MHz is typically used for TV and

radar transmission.

21. (a) 5cos 5sin2

V t tπ

ω ω = = +

⇒ 2sinI tω=

∴ 2

πφ = ⇒ cos 0

2rms rsmP E Iπ

= = W

22. (d) Density 2

m

r Lρ

π=

∴ 100 2 100m r L

m r L

ρρ

∆ ∆ ∆ ∆ × = + + ×

After substituting the values, we get the maximum

percentage error in density = 4%

23. (c) L = 40 m, v = 1080 km,

h–1 = 300 m 1sec− and B = 48.3 10 T−×

⇒ 48.3 10 40 300 1e Blv −= = × × × = V

24. (a) Relative velocity of image ... trw object

= 3 – (–3) = 6 m/sec

25. (b) Angular momentum of satellite about center of earth

remains constant, i.e., mvr = constant

⇒ 2 1

1 2

1

2

v rv

v r∝ ⇒ =

Speed is maximum at position 2,

∴ ( )( )

max 1

min 2

1

1

av r

v r a

εε

+= =

−1 1 0.0167

1.0331 1 0.0167

εε

+ += = =

− −

3m/sec –3m/sec

I O

Physics 42


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

a d a c a,b,d b,d 4 8 9 8

11. 12. 13. 14. 15. 16. 17. 18.

8 4 2 4 a,d a,c b d

1. (a)e 1 2

1 1 1

R q R R= +

⇒ 2 2 2

e 1 2

.....R q

V V VP

R R= = + +

15 40 5 100 80 5 1 1000 2500w= × + × + × + × = W ⇒ P VI=

⇒ 2500 220P I= = ⇒ 250 125

22 11I = =

2. (d) d

i

Aneυ = ∴ d

ine

Aυ= ⋅

⇒ V V

iR

Aρ

= =ℓ

∴ Vi

A

ρ =ℓ

⇒ d

V

neρ

υ=

⋅ ℓ

⇒ 4 28 19

5

2.5 10 8 10 1.6 10 0.1ρ − −=

× × × × × ×

4 5

4

2 110 1.6 10 m

8 1.6 10 6.4− −

−= = × = × Ω× ×

3. (a)1

g

g

i GS

i=

−Here 310gi A−= A

⇒ 210 ,G = Ω 10I A= ⇒ 210S −= Ω

4. (c) For conductor (Cu) resistance increases linearly and

for semiconductor resistance decreases Exponentially in

given temperature range.

5. (a, b, d) 1 1 21

1 3

( )+=

+

R V VV

R R

⇒ 1 3 2 1=V R V R

⇒ ⇒ ⇒ ⇒ 3 1 22

1 3

( )+=

+

R V VV

R R ⇒ 2 1 2 3=V R V R

6. (b, d)2 2 2

1 24/ 2 / 8

= =V V V

H t tR R R

⇒ 1 2 min.=t ⇒ 2 0.5 min.=t

7. (4) Refer Fig., The magnetic field induction at the

centroid O due to current I through one side BC of the

triangle will be 01 1 2

4

IB

r

µπ

= +. (sin θ1 + sin θ2)

It will be acting perpendicular to the plane of triangle

upwards. Total magnetic field induction at O due to

current through all the three sides of the triangle will be

01 1 2

33 sin sin

4

IB B

r

µπ

= = +[sin θ1 + sin θ2]

Here, I = 1 A, θ1 = 60° = θ2

And 2/ 2 4.5 10

mtan 60 3 2 3 2 3

BD a ar OD

−×= = = = =

°

∴ ( )

[ ]7

2

13 10 sin 60 sin 60

4.5 10 / 2 3B −

−= × × × ° + °

×

On solving, B = 4 × 10–5 T.

8. (8) Here, Am = 4 V ; Ac = 5 V

Percentage of modulation,

A 4

100 100 80% 10%.A 5

m

c

aµ = × = × = = × ⇒ a = 8

9. (9) Here, 4.5kHz; 3.45MHz 3450kHzs cv v= = =

Bandwidth 2 2 4.5 9.0kHz.sv= = × =

10. (8) Here, 4MHz, 90cv i= = °

Maximum usuable frequency seccv i=

4 sec60 4 2 8MHz= × ° = × =

11. (8) ' 2 ' 3 3 2d h R d hR= = =

or ' 9 9 100 900h h m= = × =

Increase in height of tower 2900 100 800 10a= − = = ×

⇒ 8a =

12. (4) Microwave frequency used in telephone link = 10 GHz

= 10 × 109 Hz = 1010 Hz

Frequency available for microwave communication

2%= of 10 GHz = 10 8210 2 10 Hz

100× = ×

Bandwidth of each telephone channel = 8 kHz

38 10 Hz= ×

Number of microwave telephone channels

8

4

3

2 102.5 10 2.5 10

8 10a×

= = × = ××

∴ a = 4

B I

a D

r 60° 60°

I

I

A

C

O

43 Mock Test-3

F

R

sin 30°R

30°

F

F

a

a

a a

13. (2) Iατ = ⇒ 3 sin 30 Iα° =FR

⇒ 2

2=MR

I

⇒ α 2=

⇒ 0 αtω ω= + ⇒ 2 /ω = rad s

14. (4) Since net torque about centre of rotation is zero, so we

can apply conservation of angular momentum of the

system about centre of disc =i fL L

0 2 ( / 2);ω= +I mv r comparing magnitude

∴ 0.45 0.5 0.5 0.5

0.05 9 22 2

ω× × = × × ×

∴ 4ω =

15. (a, d) 1 2=I I

⇒ 1 1 2 2=neAv neA v ⇒ 1 1 1 2 2 2=d wv d w v

Now, potential difference developed across MK

=V Bvw

⇒ 1 1 1 2

2 2 2 1

= =V v w d

V v w dand hence, correct choice is a & d.

16. (a, c) As 1 2=I I 1 1 1 1 2 2 2 2=n wd v n w d v

Now, 2 2 2 2 2 2 1 1 1 2 1

1 2 1 1 1 1 2 2 2 1 2

= = =

V B v w B w n wd B n

V B v w Bw n w d Bn

∴ Correct options are a & c.

17. (b) /0 (1 )tQ Q e τ−= −

/(1 )tQ CV e τ−= − after time interval 2 .τ

18. (d) 0 cosq Q tω=

⇒ 0 sindq

i Q tdt

ω ω= − =

⇒ kax

Ci C V V

Lω= =


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

d c a b d a 9 6 2 2/3

11. 12. 13. 14. 15. 16. 17. 18.

7 3 1 9 c a c d

1. (d) According to Lenz’s Law, current will be in

anticlockwise sense as magnetic field is increasing into

the plane of paper.

2. (c) 2 sin2

dT BiRd

θθ= (for θ small)

⇒ Td BiRdθ θ=

⇒ 2

BiLT BiR

π= =

3. (a)0

2 ( )

b

a

INdr

r b a

µ−∫ 0 0 ln

2( ) 2( )

b

a

IN INdr b

b a r b a a

µ µ = = − − ∫

4. (b) =M IA

⇒

2

224

2

π = × +

a

A a

2

224

π= × +

aa

22 21

2 2

ππ

= + +

aa a

∴ 2 ˆ12

π = +

M a Ik

5. (d) Let current be I.

Current density 22

4π

=

−

IJ

RR

⇒ 2 2

4

3 3

4

ππ= =

I IJ

R R

Field for 02

≤ =

Rr B For

2< ≤

Rr R

Applying Ampere’s Law .2πB r

2 2

2 200 2 2

44.

3 4 3 4

µµ π

π

= − = −

II R Rr r

R R

2

02

2

3 4

µπ

= −

I RB r

R r ⇒ At , 0

2= =R

r B

∴ At 2

=R

r filed is continuous.

From the above expression as r increases B increases.

For ≥r R 0.2π µ=B r I

⇒ 0

2

µπ

=I

Br

1∴ ∝B

r

⇒ At ,=r R 0

2

µπ

=I

BR

From inside expression at =r R

⇒ 2

0 02

2.

3 4 2

µ µπ π

= − =

I IRB R

R R R

This proves the continuity in the graph at .=r R From the

above only correct option is (d).

cos( / 2)T dθ cos( / 2)T dθTT

2T sin( / 2)dθ

Amperian

loop

R/2

R r

Physics 44

B A

C 2m

m

a 2a

6. (a) First we find the mutual inductance of the assembly.

Let current I flow through the larger loop.

The strength of induction at the smaller loop

ddR

IR/322

20

)(2 +=

µ

∴ Flux through smaller loop 2/322

220

)(2 dR

rIR

+=

πµ

∴ Mutual inductance 2/322

220

)(2 dR

rR

+=

µ

∴ Flux linked through coil idR

rR.

)(2 2/322

220

+=

πµ

2/322

627

)1025.2104(2

210910104−−

−−−

×+×

××××××=

ππ

101.9)10(625.15

1810

2

16 11153 −−+ ×=××

×= weber

7. (9) 121265 B C β γ− −→ + +

12 12 * 25 6( )Q m B m C c= − 12 12 2

5 6[ ( )]m B m C m C= − + ∆

12 12 2 25 6( )m B m C C mC= − − ∆ 0.014 931 4.041 9= × − =

8. (6) 2

total 1 12

B 2 2

1L

L m r

m r= +

9. (2) Assuming that masses of the two spheres are

concentrated at their centres A and B and taking x = 0 at A,

Fig., we get

0 2 3

2( 2 )cm

m m ax a AC

m m

× + ×= = =

+

2

2AC a

CB aCB a

= ∴ = = ⋅

10. (2/3) As shown in Fig., mass of small element of length dx

is 0 dm dx x dxλ λ= =

⇒ 0

0 0

0

0

( )L L

cm L

xdm x xdx

xdm

xdx

λ

λ= =

∫ ∫

∫ ∫

3 3

02 2

0

[ / 3] / 3 2

[ / 2] / 2 3

L

cm L

x L Lx

x L= = =

If 1 2

2 2,

3 3 3

L L Lx x L= = − = ⇒ 1

2

2 / 3 1

/ 3 2

x L

x L= = ⋅

11. (7) Here, m = 5 kg; 2

1 ; ?2

r m I= = =

Moment of inertia of the sphere about a tangent to the

sphere, 2 2 27 75 1 7kgm

5 5I mr= = × × = ⋅

12. (3) Here, l = 1 m, m = 0.5 kg, 6 rad/secω =

2 2

2 2 21 1 1 0 5 16 3 J

2 2 3 2 3

mlKE Iω ω

⋅ ×= = = × × = ⋅

13. (1) Here, 1 1 1,L I ω= When 2 12 ,ω ω=

and 2 2

2 2 1 1

1 12

2 2KE I Iω ω = =

2 22 1 1 1 2 1

1 1 1(2 ) 2 ;

2 2 2I I I Iω ω= × =

∴ 2 2 2 1 1 1 1 1

12

2L I I I Lω ω ω= = × = = ⇒ 2

1

1L

xL

= = ⋅

14. (9) Let 1l be the length of closed pipe 2 36cm=l = length

of open pipe

Fundamental frequency of closed pipe 1

14n

l

υ=

Frequency of first overtone of open pipe,

2

1 2

22 36

nl l

υ υ υ= × = =

As 1 2n n=

∴ 1

1

;4 364 36

ll

υ υ= =

⇒ 1

369cm.

4l = =

15. (c) A-4; B-1, 4; C- 4; D-2

A. 2 2 2 2 13[ ] [ ] [ ]

2KE KT ML T K K K ML T K− − −′ ′= ⇒ = ⇒ =

B. 2 1 1 16 [ ] [ ] [ ] [ ]F rv MLT L LT ML Tπη η η− − − −= ⇒ = ⇒

C. 2 2 2 1[ ] [ ][ ]

hE hf ML T h ML T

T− −= ⇒ = ⇒ =

D. 2 2 2( ) [ ] [ ] [ ]

[ ] [ ]

dQ K A T ML T k L K

dt x T L

−′ ′∆= ⇒ =

∆

⇒ 3 1[ ]K MLT K− −′ =

16. (a) A-4; B- 1,4; C- 4; D- 2

17. (c) c. A-1; B-4; C-2; D-3

When coin is dropped if falls is straight line where direction is same but magnitude changes with change in position. When a coin is rotated in circular path direction goes on changing but radius remains constant. In projectile motion both direction and magnitude of r changes. A coin held in your hand has constant position vector and hence it is invariant.

18. (d) A-1, B-2,4, C-3, D-4

I

R

r

L

x dm

dx

45 Mock Test-4



1. Which of the following sets have different dimensions?

a. Pressure, Young’s modulus, Stress

b. Emf, Potential difference, Electric potential

c. Heat, Work done, Energy

d. Dipole moment, Electric flux, Electric field

2. The 100 coplanar forces each equal to 10 N act on a body.

Each force makes angle 50

πwith the preceding force.

What is the resultant of the forces?

a. 1000 N b. 500 N

c. 250 N d. Zero

3 A string of negligible mass going over a clamped pulley

of mass m supports a block of mass M as shown in the

figure. The force on the pulley by the clamp is given by

a. 2Mg b. 2mg

c. 2 2( )M m m g+ + d. 2 2( ( ) )M m M g+ +

4. A simple pendulum is oscillating without damping. When

the displacement of the bob is less than maximum, its

acceleration vector a

is correctly shown in

a. b.

c. d.

5. Four holes of radius R are cut form of a thin square plate

of side 4R and mass M. The moment of inertia of the

remaining portion about z-axis is:

a. 2

12MR

π b. 24

3 4MR

π −

c. 28 10

3 16MR

π −

d. 24

3 6MR

π −

6. The radius of earth is 6400 km and g = 10 m/s2. In order

that a body of 5 kg weighs zero at the equator, the angular

speed of earth in radian/sec is:

a. 1

80 b.

1

400

c. 1

800 d.

1

1600

7. The lower end of a capillary tube of radius r is placed

vertically in water. Then with the rise of water in the

capillary, heat evolved is:

a. 2 2 2r h

dgJ

π+ b.

2 2 2

2

r h dg

J

π+

c. 2 2

2

r h dg

J

π− d.

2 2r h dg

J

π−

8. A wire of length L and cross-sectional area A is made of a

material of Young’s modulus Y. If the wire is stretched by

the amount x, the work done is:

a. 2

2

YAx

L b.

2YAx

L

c. 2

YAx

L d. 2

Yax L

9. Which of the following graphs correctly represent the

variation of β =/

= −dV dp

V with p for an ideal gas at

constant temperature?

a. b.

c. d.

10. Two rods of different materials and identical

cross-sectional area are joined face to face at one end and

their free ends are fixed to the rigid walls. If the temperature

of the surroundings is increased by 30°C, the magnitude of

the displacement of the joint of rods is: (length of the

p p

p p

y

x

a

a

a

a

m

M

Physics46

rods unit, ratio of their Young’s

modulii, ; coefficients of linear expansion are α1

and α2)

a. 2 15( )aα − b. 1 210( )aα −

c. 2 110( 2 )aα − d. 1 25(2 )aα −

11 A steel ball of mass kg moving with velocity

50 ms–1

collides with another stationary steel ball of mass

kg. During the collision, their internal

energies change equally. If T1 and T2 are rise in

temperature of balls m1 and m2

respectively, specific heat

of steel = 0.105 and mechanical equivalent of heat J = 4.2

J/cal, then:

a. 3 3

1 27.0 10 C, 1.4 10 C,T T= × ° = × °

b. 3 3

1 21.4 10 C, 7.0 10 C,T T= × ° = × °

c. 1 21.4 C, 7.0 C,T T= ° = °

d. 1 27.0 C, 1.4 C,T T= ° = °

12. Three identical dipoles are arranged as shown below.

What will be the net electric field at P ?

a. b.

c. Zero d.

13. The area of the plates of a parallel plate capacitor is A and

the distance between the plates is 10 mm. There are two

dielectric sheets in it, one of dielectric constant 10 and

thickness 6 mm and the other of dielectric constant 5 and

thickness 4 mm. The capacity of the capacitor is:

a. A b. A

c. A d. A

14. If a magnet is suspended at an angle 30o to the magnetic

meridian, it makes an angle of 45o

with the horizontal. The

real dip is

a. b.

c. d.

15. Two straight long conductors AOB and COD are

perpendicular to each other and carry currents and

The magnitude of the magnetic induction at a point P at a

distance a from the point O in a direction perpendicular to

the plane ACBD is

a. b.

c. d.

16. A circular coil and a bar magnet placed nearby are made

to move in the same direction. The coil covers a distance

of 1 m in 0.5 sec and the magnet a distance of 2 m in

1 sec. The induced e.m.f. produced in the coil:

a. Zero

b. 1 V

c. 0.5 V

d. Cannot be determined from the given information

17. Following figure shows an ac generator connected to a

"box" through a pair of terminals. The box contains

possible R, L, C or their combination, whose elements and

arrangements are not known to us. Measurements outside

the box reveals that 75 sin(sin )e tω= volt, i = 1.5 sin

(ωt + 45°) amp, then, the wrong statement is

a. There must be a capacitor in the box

b. There must be an inductor in the box

c. There must be a resistance in the box

d. The power factor is 0.707

18. The focal length of a concave mirror is f and the distance

from the object to the principle focus is x. The ratio of the

size of the image to the size of the object is

a. b.

c. d.

1 21l l= =

1 2/ 2Y Y =

1 1m =

2 0.200m =

0

1

4k

πε

=

3

.k p

x 3

2kp

x

3

2 kp

x

0

12

35ε 0

2

3ε

0

5000

7ε 01500ε

1tan ( 3 / 2)− 1tan ( 3)−

1tan ( 3 / 2)− 1tan (2 / 3)−

1i 2.i

01 2( )

2i i

a

µπ

+ 01 2( )

2i i

a

µπ

−

2 2 1/ 201 2( )

2I I

a

µπ

+ 0 1 2

1 22 ( )

I I

a I I

µπ +

f x

f

+ f

x

f

x

2

2

f

x

? ~

y

x

A B

C D

x

x

P

+Q – Q

– Q +Q

+Q – Q

p →

p →

p →

x

47 Mock Test-4

D S

r

19. The energy of incident photon is 12.375 eV while the

energy of scattered photon is 9.375 eV. Then the kinetic

energy of recoil electron is

a. 3 eV b. less than 3 eV

c. more than 3 eV d. 21.75 eV

20. In a transistor circuit shown here the base current is

35 µA. The value of the resistor Rb is

a. 123.5 kΩ b. 257 k Ω

c. 380.05 k Ω d. None of these


21. A wooden block of mass 10 g is dropped from the top of a

cliff 100 m high. Simultaneously a bullet of mass 10 g is

fired from the foot of the cliff upwards with a velocity of

100 m/s. The bullet and the wooden block will meet, each

other, after a time:

a. 10 s b. 0.5 s

c. 1 s d. 7 s

22. The displacement of a particle varies according to the

relation 4(cos sin ).x t t= +π π The amplitude of the

particle is:

a. 8 b. – 4

c. 4 d. 4 5

23. A sound wave of wavelength 32 cm enters the tube at S as

shown in the figure. Then the smallest radius r so that a

minimum of sound is heard at detector D is.

a. 7cm b. 14 cm

c.21cm d. 28 cm

24. Three rods made of the same material and having the

same cross-section have been joined as shown in the

figure. Each rod is of the same length. The left and right

ends are kept at 0°C and 90ºC respectively. The

temperature of junction of the three rods will be

a. 45°C b. 60°C

c. 30° C d. 20° C

25. Six resistance of 6 ohm each are connected to form a

hexagon. The resistance between any two adjacent

terminals is:

a. 6 Ω b. 36 Ω

c. 5 Ω d. 1 ohm

0°C

90°C

90°C

E C

B

Rb RL

9V


Physics 48

JEE ADVANCE PAPER-I



1. A glass tube of uniform internal radius r has a valve

separating the two identical ends. Initially, the valve is in

a tightly closed position. End 1 has a hemispherical soap

bubble of radius r. End 2 has sub-hemispherical soap

bubble as shown in figure. Just after opening the valve,

a. air from end 1 flows towards end 2. No change in the

volume of the soap bubbles

b. air from end 1 flows towards end 2. Volume of the soap

bubble at end 1 decreases

c. no changes occurs

d. air from end 2 flows towards end 1. volume of the soap

bubble at end 1 increases

2. A thin uniform cylinder shell, closed at both ends is

partially filled with water. It is floating vertically in water

in half-submerged state. If ρc is the relative density of the

material of the shell with respect to water, then the correct

statement is that the shell is.

a. more than half-filled if ρc is less than 0.5

b. more than half-filled if ρc is less than 1.0

c. half-filled if ρc is more than 0.5

d. less than half- filled if ρc is less than 0.5

3. An open glass tube is immersed in mercury in such a way

that a length of 8 cm extends above the mercury level. The

open end of the tube is then closed and sealed and the tube

is raised vertically up by additional 46 cm. What will be

length of the air column above mercury in the tube now?

(Atmospheric pressure = 76 cm of Hg)

a. 38 cm b. 6 cm

c. 16 cm d. 22 cm

4. There is a circular tube in a vertical plane. Two liquids

which do not mix and of densities d1 and d2 are filled in

the tube. Each liquid subtends 90° angle at centre. Radius

joining their interface makes an angle α with vertical.

Ratio 1

2

d

d is

a. 1 tan

1 tan

αα

+−

b. 1 sin

1 cos

αα

+−

c. 1 sin

1 sin

αα

+−

d. 1 cos

1 cos

αα

+−

5. On heating water, bubbles being formed at the bottom of

the vessel detaches and rise. Take the bubbles to be

spheres of radius R and making a circular contact of radius

r with the bottom of the vessel. If r << R, and the surface

tension of water is T, value of r just before bubbles

detaches is: (density of water is wρ )

a. 2 wgRT

ρ

b. 2 3 wgRT

ρ

c. 2 2

3wgRT

ρ d. 2

6wgRT

ρ

6. Two solid spheres A and B of equal volumes but of

different densities Ad and

Bd are connected by a string.

They are fully immersed in a fluid of density .Fd They

get arranged into an equilibrium state as shown in the

figure with a tension in the string. The arrangement is

possible only if

a. A Fd d< b.

B Fd d>

c. A Fd d> d. 2A B Fd d d+ =

A

B

R

2r

d

α

d

2 1

49 Mock Test-4



7. A 25 watt bulb, which is producing monochromatic light

of wavelength 6600 Å is used to illuminated a metal

surface. If the surface has 3% efficiency for photoelectric

effect, then the photoelectric current produced in deci

ampere is (use 346.6 10 Js) :h−= ×

8. de-Broglie wavelength associated with an electron

accelerated through a potential difference 4 V is 1.λ When

accelerating voltage is decreased by 3 V, its de-Broglie

wavelength is 2.λ The ratio 1

2

λλ

is

9. What is the energy in eV that should be added to an

electron of energy 2 eV to reduce its de-Broglie

wavelength from 1 nm to 0.5 nm?

10. The angular momentum of electron in 2nd orbit of

hydrogen atom is L. The angular momentum of electron in

4th orbit of hydrogen atom is n L where n = ?

11. For an atom of an ion having single electron, the

wavelength observed 1 2λ = are units and 3 3λ = units

figure. The value of missing wavelength 2λ is

12. How many different wavelength may be observed in the

spectrum from a hydrogen sample if the atoms are excited

to 3rd excited state?

13. Light waves from two coherent sources having intensities

I and 2I cross each other at a point with a phase diff. of

60º. What is the resultant intensity at the point?

14. A pn-junction diode can withstand currents upto 10 mA.

When it is forward biased, the potential drop across it is

1.0 V. Assuming that this potential drop is independent of

the current, find the maximum voltage of the battery used

to forward bias the diode when a resistance of is

connected in series with the diode.





The general motion of a rigid body can be considered to be a

combination of (i) a motion of its centre of mass about an axis,

and (ii) its motion about an instantaneous axis passing through

the centre of mass. These axes need not be stationary. Consider,

for example, a thin uniform disc welded (rigidly fixed)

horizontally at its rim to a massless stick, as shown in the

figure.

When the disc-stick system is rotated about the origin on a

horizontal frictionless plane with angular speed ,ω the motion

at any instant can be taken as a combination of (i) a rotation of

the centre of mass of the disc about the z-axis, and (ii) a

rotating of the disc through an instantaneous vertical axis

passing through its centre of mass (as is seen from the changed

orientation of points P and Q). Both these motions have the

same angular speed ω in this case.

Now consider two similar systems as shown in the figure: Case

(a) the disc with its face vertical and parallel to x-z plane; Case

(b) the disc with its face making an angle of 45° with x-y plane

and its horizontal diameter parallel to x-axis. In both the cases,

the disc is welded at point P, and the systems are rotated with

constant angular speed ω about the z-axis.

15. Which of the following statements regarding the angular

speed about the instantaneous axis (passing through the

centre of mass) is correct?

a. It is 2ω for both the cases

b. It is ω for case a.; and 2

ωfor case b.

c. It is ω for case a.; and ω for case b.

d. It is ω for both the cases

z ω

Q

y

x

P

z ω Q

y

x

P 45°

ω

P Q y P

Q

x

1λ

n3 orbit

n2 orbit

n1 orbit

3λ

2λ

Physics 50

16. Which of the following statements about the instantaneous

axis (passing through the centre of mass) is correct?

a. It is vertical for both the cases a. and b.

b. It is vertical for case a.; and is at 45°to the x-z plane

and lies in the plane of the disc for case b.

c. It is horizontal for case a.; and is at 45° to the x-z plane

and is normal to the plane of the disc for case b.

d. It is vertical for case a.; and is at 45° to the x-z plane

and is normal to the plane of the disc for case b.


A frame of reference that is accelerated with respect to an

inertial frame of reference is called a non-inertial frame of

reference. A coordinate system fixed on a circular disc rotating

about a fixed axis with a constant angular velocity ω is an

example of a non-inertial frame of reference.

The relationship between the force rotF

experienced by a

particle of mass m moving on the rotating disc and the force

inF

experienced by the particle in an inertial frame of reference

is, rot in rot2 ( ) ( ) ,F F m v m rω ω ω= + × + × ×

Where rotv

is the velocity of the particle in the rotating frame of

reference and r

is the position vector of the particle with

respect to the centre of the disc.

Now consider a smooth slot along a diameter of a disc of radius

R rotating counter-clockwise with a constant angular speed ω

about its vertical axis through its center. We assign a coordinate

system with the origin at the center of the disc, the x-axis along

the slot, the y-axis perpendicular to the slot and the z-axis along

the rotation axis ˆ( ).kω ω=

A small block of mass m is gently

placed in the slot at ˆ( / 2)r R i=

at 0t = and is constrained to

move only along the slot.

17. The distance r of the block at time t is:

a. cos 22

Rtω b. 2 2( )

4t tR

e eω ω−+

c. cos2

Rtω d. ( )

4t tR

e eω ω−+

18. The net reaction of the disc on the block is :

a. 2 ˆˆsinm R tj mgkω ω −

b. 2 ˆˆcos 6m R tj mgkω ω− −

c. 21 ˆˆ( )2

t tm R e e j mgkω ωω −− +

d. 2 2 21 ˆˆ( )2

t tm R e e j mgkω ωω −− +

R

R/2

m

ω


51 Mock Test-4




1. One end of a horizontal thick copper wire of length 2L

and radius 2R is welded to an end of another horizontal

thin copper wire of length L and radius R. When the

arrangement is stretched by applying forces at two ends,

the ratio of the elongation in the thin wire to that in the

thick wire is

a. 0.25 b. 0.50

c. 2.00 d. 4.00

2. The pressure that has to be applied to the ends of a steel

wire of length 10 cm to keep its length constant when its

temperature is raised by 100°C is: (For steel Young’s

modulus 2 × 1011 Nm–2 and coefficient of thermal

expansion is 1.1 × 10–5 K–1 )

a. 72.2 10 Pa× b. 62.2 10 Pa×

c. 82.2 10 Pa× d. 92.2 10 Pa×

3. A pendulum made of a uniform wire of cross-sectional

area A has time period T. When an additional mass M is

added to its bob, the time period changes to .MT If the

Young's modulus of the material of the wire is Y then 1/Y

is equal to: (g = gravitational acceleration)

a.

2

1MT A

T Mg

−

b. 2

1MT Mg

T A

−

c. 2

1 MT A

T Mg

−

d.

2

1M

T A

T Mg

−

4. Function 2 2sin cos sin cosω ω ω ω= + +x A t B t C t t

represents SHM.

a. For any value of A, B and C (except C = 0)

b. If A = – B; C = 2B, amplitude | 2 |B=

c. If A = B; C = 0

d. If A = B; C = 2B amplitude = |B|

5. Two coaxial solenoids of different radii carry current I in

the same direction. Let 1F

be the magnetic force on the

inner solenoid due to the outer one and 2F

be the magnetic

force on the outer solenoid due to the inner one. Then:

a. 1 2 0F F= =

b. 1F

is radially inwards and 2F

is radially outwards

c. 1F

is radially inwards and 2 0F =

d. 1F

is radially outwards and 2 0F =

6. Consider a spherical shell of radius R at temperature T.

The black body radiation inside it can be considered as an

ideal gas of photons with internal energy per unit volume

4Uu T

V= ∝ and pressure

1.

3

UP

V

=

If the shell now

undergoes an adiabatic expansion the relation between T

and R is:

a. RT e−∝ b. 3RT e−∝

c. 1

TR

∝ d. 3

1T

R∝



7. An electron in an excited state of Li2+ ion has angular

momentum 3 / 2 .πh The de-Broglie wavelength of the

electron in this state is pπa0 (where a0 is the Bohr radius).

The value of p is

8. Consider a hydrogen atom with its electron in the nth

orbital. An electromagnetic radiation of wavelength 90 nm

is used to ionize the atom. If the kinetic energy of the

ejected electron is 10.4 eV, then the value of n is (hc =

1242 eV nm)

9. A hydrogen atom in its ground state is irradiated by light

of wavelength 970 Å. Taking hc/ e 61.237 10 −= × eV m and

the ground state energy of hydrogen atom as –13.6 eV, the

number of lines present in the emission spectrum is:

10. To determine the half life of a radioactive element, a

student plots a graph of( )

n

dN t

dtℓ versus t.

Here dN(t)/dt

is the rate of radioactive decay at time t. If

the number of radioactive nuclei of this element decreases

by a factor of p after 4.16 years, the value of p is

Years

6

5

4

3

2

1 2 3 4 5 6 7 8

ℓn|dN

(t)/dt|

L θ

I I

Physics 52

11. The activity of a freshly prepared radioactive sample is 1010 disintegrations per second, whose mean life is 910 s.The mass of an atom of this radioisotope is 2510 kg.−

kg. The mass (in mg) of the radioactive sample is

12. A freshly prepared sample of a radioisotope of half-life

1386s has activity 103 disintegrations per second.

Given that ln 2 = 0.693, the fraction of the initial number

of nuclei (expressed in nearest integer percentage)

that will decay in the first 80s after preparation of the

sample is

13. A nuclear power plant supplying electrical power to a

village uses a radioactive material of half life T years as

the fuel. The amount of fuel at the beginning is such that

the total power requirement of the village is 12.5 % of the

electrical power available from the plant at that time. If

the plant is able to meet the total power needs of the

village for a maximum period of nT years, then the value

of n is

14. For a radioactive material, its activity A and rate of change

of its activity R are defined as = −dN

Adt

and ,= −dA

Rdt

where N(t) is the number of nuclei at time t. Two

radioactive sources P (mean lifeτ ) and Q (mean life 2 )τ

have the same activity at t = 0. Their rates of change of

activities at 2τ=t are PR and ,QR respectively. If

,=P

Q

R n

R e then the value of n is



15. There is a definite relation between velocity, mass and

acceleration of body to know about the work done by the

forces applied on body. Column I has some statements

showing some relations which are related with some

statements in Column II. Match the correct options.

Column I Column II

(A) If kinetic energy is K the

momentum P is 0

1. zero

(B) If momentum p is zero the

kinetic energy is 0 2. 2mk

(C) If different masses have

same kinetic energy the

ratio of their momenta is 0

3. 2

1

m

m

(D) If different masses have

same momentum, the ratio

of their kinetic energy is

4. 1

2

m

m

a. A-1, B-2, C-3, D-4 b. A-2, B-1, C-4, D-3

c. A-4, B-3, C-2, D-1 d. A-3, B-2, C-1, D-4

16. Waves are classified on the basis of frequency range.

Column I mentions the type of waves while column II

gives the corresponding frequency range. Can you match

the proper options?

Column I Column II

(A) Infrasonic 1. Objects having velocity

more than velocity of

sound in air at 0°C

(B) Audible 2. Frequency 20 Hz<

(C) Ultrasonics 3. Frequency 20,000 Hz>

(D) Supersonics 4. Frequency lies between

20 Hz to 20, 000 Hz

a. A-4, B-3, C-2, D-1 b. A-2, B-4, C-3, D-1

c. A-4, B-4, C-1, D-2 d. A-3, B-2, C-1, D-4

17. Various electromagnetic waves are given in column I and

various frequency ranges in column II

Column I Column II

(A) Radiowaves 1. 16 211 10 to 3×10 Hz×

(B) γ-rays 2. 9 111 10 to 3×10 Hz×

(C) Microwaves 3. 18 223 10 to 5×10 Hz×

(D) X-rays 4. 5 95 10 to 10 Hz×

a. A-1, B-2, C-3, D-4 b. A-2, B-3, C-4, D-1

c. A-3, B-4, C-1, D-2 d. A-4, B-3, C-2, D-1

18. In each of the following questions, match column I and

column II, and select the correct match out of four given

choices.

Column I Column II

(A) Transducer 1. Range of frequencies over

which communication

system works

(B) Attenuation 2. A device that has input in

electrical form or provides

output in electrical form

(C) Range 3. Loss of strength of a

signal during propagation.

(D) Bandwidth 4. Argest distance between

transmitter and receiver.

a. A-,2 B-4, C-1, D-3 b. A-4, B-3, C-1, D-2

c. A-1, B-2, C-3, D-4 d. A-2, B-3, C-4, D-1

53 Mock Test-4

ANSWERS & SOLUTIONS

JEE-Main

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

d d d c c c b a a c

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

c c c a c a b b a b

21. 22. 23. 24. 25.

c d b b c

1. (d) Dipole moment = (charge) × (distance)

Electric flux = (electric field) × (area)

2. (d) Total angle 100 250

ππ= × =

So, all the force will pass through one point and all forces

will be balanced. i.e. their resultant will be zero.

3. (d) Free body diagram of pulley is shown in figure.

Pulley is in equilibrium under four forces. Three forces as

shown in figure and the fourth, which is equal and

opposite to the resultant of these three forces, is the force

applied by the clamp on the pulley (say F).

Resultant R of these three forces is 2 2( ( ) )R M m M g= + +

Therefore, the force F is equal and opposite to R as shown

in figure.

∴ 2 2( ( ) ) .F M m M g= + +

4. (c) Net acceleration a

of the bob in position

B has two components.

(a) na =

radial acceleration (towards BA)

(b) ta =

tangential acceleration (perpendicular to BA)

Therefore, direction of a

is correctly shown in option (c).

5. (c) M = Mass of the square plate before cutting the holes

Mass of one hole 2

216 16

M Mm R

R

ππ

= =

∴ Moment of inertia of the remaining portion,

4square holeI I I= −

( ) ( )2

2 2 216 16 4 212 2

M mRR R m R

= + − +

2 2810

3MR mR= − 28 10

3 16MR

π = −

6. (c) 2g g Rω′ = − (at equator)

Given 20g g Rω′ = ⇒ =

⇒ 3

10 1

6400 10 800

g

Rω = = =

×rad/s

7. (b) When the tube is placed vertically in water, water rises

through height h given by 2 cosT

hrdg

=θ

Upward force 2 cosr T= ×π θ

Work done by this force in raising water column through

height h is given by (2 cosW rT h=∆ π θ)

2 2(2 cos (2 cos

2cos

rhdgrh T = rh r h dg

= = π θ) π θ) π

θ

However, the increase in potential energy pE∆ of the raised

water column ,2

hmg= where m is the mass of the raised

column of water.

∴ 2m r hdπ=

So, 2 2

2( )2 2

p

gh r h dgE r hd

ππ ∆ = =

Further, 2 2

2p

r h dgW E

π∆ − ∆ =

The part ( )pW E∆ − ∆ is used in doing work against viscous

forces and frictional forces between water and glass

surface and appears as heat.

So, Heat released 2 2

2pW E r h dg

J J

π∆ − ∆=

8. (a) Energy stored, 1

2U = × (force)×(extension)

1

2Fx=

As xFL YAY F

Ax L= ⇒ =

∴ 21 1

2 2xYA YAx

U xL L

= =

9. (a) β =/dV dp

V= − = compressibility of gas

1

Bulk modulus of elasticity= and β =

1

p= under isothermal

na

A

B a

ta

R

F

Mg + mg

Mg

T = Mg

mg T=Mg

Physics54

conditions. Thus, β versus p graph will be a rectangular

hyperbola.

10. (c) Applying,

temperature difference across the rods are 20°C and 10°C

respectively.

∴ Displacement of the joint

11. (c) J m1s ∆θ1

⇒ ∆θ1

2

4Js

v= = = 1.4 °C

1 2m s m s=1 2

∆θ ∆θ

⇒ 1

2

11.4 7 C

0.2

m

m= = × = °

2 1∆θ ∆θ

12. (c) Point P lies at equatorial positions of dipole 1 and 2

and axial position of dipole 3. Hence field at P

Due to dipole 1 (towards left)

Due to dipole 2 (towards left)

Due to dipole 3 (towards right)

So, Net field at P will be zero.

13. (c) The two capacitor C1 and C2

are connected in series

This gives

14. (a) Let the real dip be φ, then

For apparent dip,

tancos

V

H

B

B′ =φ

β

or φ or φ

15. (c) At P :

16. (a) Speed of the magnet m/s

Speed of the coil m/s

Relative speed between coil and magnet is zero, so there

is no induced emf in the coil.

17. (b) Since voltage is lagging behind the current, so there

must be no inductor in the box.

18. (b) where ∴

19. (a) Kinetic energy of recoil electron,

12.375 9.375 3eV= − =

20. (b) kΩ

21. (c)

⇒

⇒ ⇒ = 1s

( )1 2KAQ

l

θ θ−=

2 2 1 1t tα α= −

( )2 1 2 110 20 10 2α α α α= − = −

2

1 1 1

1 1

2 2m v J m s⋅ = ∆

( )2

3

50

4 4 4.2 10 0.105∆ = =

× × ×

1 3

.k pE

x=

2 2

.k pE

x=

3 3

.(2 )k pE

x=

1 0 2 01 2

1 2

,K A K A

C Cd d

ε ε= =

1 2

1 2 1 0 2 0

1 1 1,

d d

C C C K A K Aε ε= + =

1 2

0 1 2

1 d d

A K Kε

= +

3 3

0

1 6 10 4 10

10 5Aε

− − × ×= +

3

0

1 14 10

10Aε

−×=

0

5000

7C Aε=

tan V

H

B

Bφ =

2

cos cos30 3

V V V

H H H

B B B

B B Bβ= = =

°

2tan 45 .tan

3° = 1 3

tan2

φ − =

2 2

1 2netB B B= +

2 2

0 01 22 2

4 4

i i

a a

µ µπ π

= +

2 2 1/ 20

1 2( )2

i ia

µπ

= +

1

22 /

1v m s= =

2

12 /

0.5v m s= =

;I f

O f u=

−u f x= + I f

O x= −

kE hv hv′= −

6

9257

35 10b b b b

V i R R k−

= ⇒ = = Ω×

1 2 100s s+ =

2 21 1100

2 2gt ut gt m+ − =

100ut m=100 100

100t s

u= = =

S N

v1

v2

i2

D

O

B

C

A

i1

a

B1

B2

P

P

+Q – Q

– Q +Q

+Q – Q

E1 E2 E3

1

3

2

10 mm

6 mm 4 mm

55Mock Test-4

22. (d) 4(cos sin )x t t= +π π

Standard equation of displacement is

Comparing the given equation with standard equation

23. (b) Path difference 162

32

2)2( ===−

λπ rr

⇒ 16

14cm2

rπ

= =−

24. (b) Let θ be the temperature of the junction (say B).

Thermal resistance of all the three rods is equal. Rate of

heat flow through AB + Rate of heat flow through CB =

Rate of heat flow through BD

∴ 90 90 0

R R R

°− °− −+ =

θ θ θ, Here, R =Thermal resistance

∴ 3 180 ,= °θ or 60 C= °θ

25. (c) Resistance and are in parallel; so

equivalent resistance


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

b a a a c a,b,d 4 2 6 2

11. 12. 13. 14. 15. 16. 17. 18.

6 6 4 2 d a d c

1. (b) P1 = pressure just inside the bubble at the end

0

42

TP

R= +

2. (a) If water is half filled shell displaced( )ρ= ∆M g V g

⇒ shell( )2

∆ =

VV

⇒ liwuid shellshell

2

ρ=

VM

⇒

liquidshell

shell 2

ρ =

M

V

⇒ liquid

shell2

ρσ =

⇒

1. .

2=R d

1If .

2<R d , Then the cylinder should be more than half

filled so that it is half submerged and floating.

3. (a) 0(54 ) 8P h P′ − = (Boyel’s Law)

⇒ 0 8

54

PP

h′ =

− ⇒ 0P gh Pρ′ + =

⇒ 00

8

54

Pgh P

hρ+ =

−

⇒ 0

46

54

hgh P

hρ

− = − 0( (76))P gρ=

⇒ 46

(76)54

hh

h

− = −

⇒ 254 76 46 76h h h− = × −

⇒ 2 130 76 46 0h h− + × = ⇒ ( 38)( 92) 0h h− − =

⇒ ⇒ 38cmh=

4. (a) Pressure at 0 level is same from both sides.

1(1 sin )d α−

1 2(1 cos ) [cos sin ]d dα α α= − + +

⇒ 1

2

sin cos

cos sin

d

d

α αα α

+=

−

5. (c) 2 2

3

wgR

T

ρ, The downward force on the bubble due to

surface tension 2 . sinr T= π θ

22 Tr

R

π=

1 14 2 cos sin

2 2t tπ π

+

4 2 sin cos cos sin2 2

t tπ π

π π = + 4 2 sin

4t

ππ = +

( )sinx a tω φ= +

4 2a =

( )6 5 30× = Ω 6 Ω

30 65

30 6R

×= = Ω

+

R

r

θ

θ δT

ααα

d

0

(54-h)

54 cm

h

P’ P0

0°C

90°C

90°C

A

B

C

D

2l

2Y θ Y θ1 θ2

l

Not possible

38,92h =↓

Physics 56

The upward buoyant force exceeds, the surface tension

force then the bubble detaches.

∴ 2

34 2

3w

TrR g

R

ππ ρ = ⇒ 2 2

3wgr RT

ρ=

6. (a, b, d)

7. (4) No. of photons emitted per second by 25 watt source

will be

10

34 8 19

25 (6600 10 ) 25

/ (6.6 10 ) (3 10 ) 3 10

E En

hc hc

λλ

−

− −

× ×= = = =

× × × ×

Current, I = 3% of number of photoelectrons emitted per

sec × charge of electron

19

19

3 3 251.6 10 0.4A 4 deciampere

100 100 3 10ne

−−= × × × = =

×

8. (2) de-Broglie wavelength, 4 1

or2 V Vmq

λ λ= ∝

∴ 2 1

1 2

V 4V2

V 4V 3V

λλ

= = =−

9. (6) As 2 K

h

mEλ =

or 2

2 2

1or

2K K

hE E

mλ λ= ∝

∴ 1

2

2 2

2

1

0.5 1

1 4K

K

E

E

λλ

= = =

or 1 1

4 4 2 8K KE E eV eV= = × == 8 eV

Increase in energy 2 1

8 2 6eVK KE E= − = − =

10. (2) 22

hL

π =

and ' 42

hL

π =

∴ ' 2 ,L L n L= = where n = 2

11. (6) As is clear from figure. 1 2 3E E E= +

1 2 3

hc hc hc

λ λ λ= +

∴ 2 1 3

1 1 1 1 1 1

2 3 6λ λ λ= − = − = = 6 units

12. (6) 3rd excited state means n = 4. No. of different

wavelength observed ( 1) 4(4 1)

62 2

n nN

− −= = =

13. (4) Here, 1I I=

2 2 , 60I I φ= = °

Amplitude R of resultant wave is

2 2 2 cosR a b ab φ= + +

2 2 2 2 cosR a b ab φ= + + As intensity ∝ (amplitude)2

∴ Resultant intensity, 1 2 1 22 cosRI I I I I φ= + +

2 2 2 cos60I I I I= + + × × °

3 2 (3 2) 4.414I I I I= + = + =

14. (2) See figure

Voltage drop across diode is

dV V= 1.0V

Voltage drop across R is RV RI=

3100 10 10−= × × = 1.0V

max 1.0 1.0 2.0d RV V V V= + = + == 2.0V

15. (d) It is ω for both case.

As angular velocity of a rigid body about any point should

be same. = +

PC COv v v

⇒ 'ω= + +

PO PC COv r v

ˆˆ ˆ ˆ( ) .4522 2

d dLi k j L iω ω ω

− = − + × + − + °

⇒ ˆpr j=ℓ

⇒ ˆˆcos45 sin 452 2c

d dr j k

= + ° + °

ℓ

⇒ ˆˆcos45 sin 452 2

= − ° − °

PC

d dr j k

⇒ ˆcos452

CO

dv iω

= − + °

ℓ

⇒ ˆ ˆ( ) '2 2 2 2

ωω= − + ×d d

i i j ⇒ ˆ'ω ω=

k

16. (a) 2 2ω= +ℓV r TOPView

2 sin

2

VW

r

θ =

2 2

2 2

ω +=

+

ℓ

ℓr

r r

rω=

kω

P

ˆ( dcos 45 )iℓ− + °ω

iℓ−ω

d cos 45°

V

I

VR

R

Vd

A

B

2 Fd Vg

( )A Bd d Vg+

57 Mock Test-4

Here there is a relv of P′ and Q′ (not for P, Q) (vrel)P’Q’ will

be in horizontal plane

⇒ ω must be along vertical. Same situation will be there for

P as mentioned ω should be same for both cases.

17. (d)2

mr maω =

⇒ 2a rω= ⇒ 2vdv

rdr

ω=

⇒ 2

0 / 2

v r

R

vdv rdrω=∫ ∫

⇒2

2

4

Rv rω= −

2

/ 2 02

4

v t

R

drdt

Rr

ω=

−∫ ∫ . . .(i)

Assume: sec2

Rr = θ

⇒ sec tan2

Rdr d= θ θ θ

⇒ 2

02

sec tan2

tan4

tR

d

dtR

θ θ θ=

θ∫ ∫ω

⇒ 2 22 4r r R

t nR R

ω −

= +

ℓ

⇒ [ ]4

t tRr e eω ω−= +

18. (c) rot in rotˆ ˆ ˆˆ ˆ2 ( ) ( )F F m v i k m k ri kω ω ω= + × + × ×

2 2in rot

ˆ ˆ ˆ2 ( )mr i F mv j m riω ω ω= + − +

inˆ2 rF mv jω=

. . .(i)

4

t tRr e eω ω− = +

4

t t

r

dr Rv e e

dt

ω ωω ω − = = −

inˆ2

4t tR

F m e e jω ωωω− = −

Also reaction is due to disc surface then

2

ˆˆ2

t t

reaction

mRF e e j mgk

ω ωω − = − +


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

c c a a,b,c a a,b 2 2 6 8

11. 12. 13. 14. 15. 16. 17. 18.

1 4 3 2 b b d d

1. (c)

⇒ 2 2

1 2

4,

2

π π= =

R x R yk k

L L

⇒ 1 2= =F k x k y ⇒ 1

2

2= =ky

x k

2. (c) F

YA

∆=

ℓ

ℓ⇒ Tα

∆= ∆ℓ

ℓ

⇒ F

P Y TA

α= = ∆

11 5 8 22 10 1.1 10 100 2.2 10 /−= × × × × = ×

Pa

3. (a) 12 ,Tg

π=ℓ

22MTg

π=ℓ

2 2 14 ,T

gπ= ⋅ℓ

2 2 22MT

gπ=ℓ

⇒ 2

22

1

MT xT

=ℓ

ℓ

⇒ 2

2 2 12

1 1

1 1MT

T

−− = − =

ℓ ℓ ℓ

ℓ ℓ

⇒ 1

2 1( )

Mg

Aγ = ⋅

−ℓ

ℓ ℓ

⇒

2

2 1

1

( )11MTA A

Mg Mg Tγ

− = ⋅ = −

ℓ ℓ

ℓ

4. (a, b, c) (1 cos2 ) (1 cos 2 ) sin 22 2 2

ω ω ω= − + + +A B C

x t t t

For 0, 0= =A B sin 22

ω=C

x t

For = −A B and 2=C B cos 2 sin 2ω ω= +x B t B t

Amplitude | 2 |= B

For ; 0= =A B C =x A

Hence this is not correct option

For , 2= =A B C B sin 2ω= +x B B t

It is also represents SHM.

5. (a) Solenoid consists of circular current loops which are

placed in uniform field, so magnetic fore on both

solenoids 0.=

F F

2L, 2R L, R

2 20 rω +ℓ

2 2r+ℓ

2r+ℓ

1p

θ θ

θ

ˆ( )kω ˆ( )y j

rv

i

r

Physics 58

⇒ 1 2 0F F= =

6. (c) 41

3

UP T

V

= ∝

For an ideal gas PV nRT=

∴ nR T

PV

′= ( R′ -molar gas constant)

∴ RT e−∝ ⇒ 3nRT

V

′∝

∴ 31T

V∝ ⇒

3

3

14

3

T

Rπ

∝

∴ 3

3

1T

R∝ ⇒

1T

R∝

7. (2) 3

2 2π π= =nh h

mvr de-Broglie wavelength

2

00

(3)2 22

3 3

π πλ π= = = =

Li

ah ra

mv z

8. (2) photon ionize atom kinetic energyE E E= +

⇒ 2

1242 13.610.4

90 n= + . From this, 2n =

9. (6) [Here we are assuming that in original paper atom

⇒ atoms] 2 2

1 1 1237013.6

1 970n

− =

⇒ 2 16n = ⇒ 4n =

No. of lines 42C=

4 36

2

×= =

10. (8) 0tN N e λ−=

⇒ 0| / | ( )n dN dt n N tλ λ−ℓ ℓ

From graph, 1

2λ = per year

⇒ 1/ 2

0.6931.386

1/ 2t = = year

⇒ 1/ 24.16 yrs =3t ∴ 8p =

11. (1) A Nλ= ⇒A

N Aτλ

= =

So 25 10 9 610 10 10 10 kg 1 mgM mN mAτ − −= = = × × = =

12. (4) –(1 ) 1 1 (1 )t tf e e t t

λ λ λ λ−= − = − ≈ − − =

⇒ 0.04f = . Hence % decay 4%≈

13. (3) 0

1

2

t

TA

A

=

Where, 0A is the initial activity of the radioactive material

and A is the activity at t.

So, 12.5 1

100 2

t

T =

∴ 3 .t T=

14. (2) λP1 1

;2

λ λτ τ

= =p Q

⇒ 0

0

( ) 2, At 2

P

Q

t

PP P

t

Q QQ

A eR Rt

R R eA e

−

−= = =

λ

λ

λτ;

λ

15. (b) A-2, B-1, C-4, D-3

Kinetic energy 21

2K mv=

or,

2

2

pK

m= or 2p mK=

If 0,K = 0p = as above

For different mass 1m and 2m if 1 2E E=

2 21 2

1 22 2

p p

m m= or 1 1

2 2

p m

p m=

If 1 2p p= for different masses.2

1 1 2 2

2 1 2 1

2

2

E p m m

E m p m= × =

16. (b) A-2, B-4, C-3, D-1

Waves having frequency less than minimum audible

frequency are infrasonics.

Waves of frequency between 20Hz to 20,000Hz are called

audible waves. Waves having frequency > 20,000Hz

are ultrasonic waves. Supersonic objects are those which

have velocity more than velocity of sound.

17. (d) A-4, B-3, C-2, D-1

The frequency ranges of various waves are as under :

Radiowaves; 55 10× to 910 ; Hz;

γ-rays; 183 10× to 225 10× Hz

Microwaves; 91 10× to 113 10 ;× Hz;

X-rays; 161 10× to 213 10 .× Hz.

18. (d) A-2, B-3, C-4, D-1 A transducer is a device that has input in electrical form or

provides output in electrical form. Attenuation is loss of strength of a signal during propagation. Range is largest distance between transmitter and receiver. Bandwidth is range of frequencies over which communication system works.

F1

B = 0

B2

x x x x x x x x x x x x

x x x x x x x x x x x x

59Mock Test-5



1. Frequency is the function of density )(ρ , length )(a and

surface tension )(T . Then its value is

a. 1/ 2 3/ 2k a Tρ / b. 3/ 2 3/ 2

/k a Tρ

c. 1/ 2 3/ 2 3/ 4

/k a Tρ d. 1/ 2 1/ 2 3/ 2

/k a Tρ

2. With respect to a rectangular cartesian coordinate system,

three vectors are expressed as: ˆ ˆ ˆ ˆ4 , 3 2a i j b i j= − = − +rr

and ˆc k= −r

where ˆˆ ˆ, ,i j k are unit vectors, along the X, Y

and Z-axis respectively. The unit vectors r along the

direction of sum of these vector is:

a. 1 ˆˆ ˆˆ ( )3

r i j k= + − b. 1 ˆˆ ˆˆ ( )2

r i j k= + −

c. 1 ˆˆ ˆˆ ( )3

r i j k= − + d. 1 ˆˆ ˆˆ ( )2

r i j k= + +

3. If and represent the work done in moving a

particle from A to B along three different paths 1, 2 and 3

respectively (as shown) in the gravitational field of a

point mass m. Find the correct relation between

and

a. W1 > W2 > W3 b. W1 = W2 = W3

c. W1 < W2 < W3 d. W2 > W1 > W3

4. The moment of inertia of a uniform rod of length 2l and

mass m about an axis passing through its center and

inclined at an angle is:

a. b.

c. d.

5. A mass M is suspended from a spring of negligible mass.

The spring is pulled a little and then released, so that the

mass executes SHM of time period T. If the mass is

increased by m, the time period becomes 5T/3. The ratio

of m/M is

a. b. c. d.

6. The gravitational potential difference between the surface

of a planet and a point 10 m above is 4.0 J/kg. The

gravitational field in this region, assumed uniform is

a. 0.025 N/kg b. 0.40 N/kg

c. 40 N/kg d. 4.0 N/kg

7. 1000 drops of water all of same size join together to form

a single drop and the energy released raises the

temperature of the drop. Given the T is the surface tension

of water. r the radius of each small drop, the density of

liquid, J the mechanical equivalent of heat. What is the

rise in temperature?

a. b.

c. d. None of these

8. A simple pendulum is made by attaching a 1 kg bob to

5 m long copper wire. Its period is T. now if 1 kg bob is

replaced by 10 kg bob, the period of oscillations:

a. remains T

b. becomes greater than T

c. becomes less than T

d. any of above answer depends on locality

9. On which of the given scales of temperature, the temperature

is inverse negative?

a. Celsius b. Fahrenheit

c. Reaumur d. Kelvin

10. An ideal gas is initially at temperature T and volume V. Its

volume is increased by due to an increase in

temperature pressure remaining constant. The quantity

varies with temperature is:

a. b.

c. d.

1 2,W W 3W

1 2,W W

3.W

xx′

α

22sin

3

mlα

22sin

12

mlα

22cos

6

mlα

22cos

2

mlα

5

3

3

5

16

9

28

9

ρ

/T Jr 10 /T Jr

100 /T Jr

∆V

,∆T

/δ = ∆ ∆V V T

T T

T T

T T

T T

A B α

C

x′

x

A

B

1 2 3

Physics 60

11. A lead ball moving with velocity V strikes a wall and

stops. If 50T of its energy is converted into heat, then

what will be the increases in temperature (specific heat of

lead is s)?

a. 22V

Js b.

2

4

V

Js

c. 2V s

J d.

2

2

V s

J

12. The plots of intensity versus wavelength for three black

bodies at temperatures 1 2,T T and 3T respectively are as

shown. Their temperatures are such that

a. 1 2 3> >T T T b.

1 3 2> >T T T

c. 2 3 1> >T T T d.

3 2 1> >T T T

13. The electric field in a region surrounding the origin is

uniform and along the x-axis. A small circle is drawn with

the centre at the origin cutting the axes at points A, B, C, D

having co-ordinates (a, 0), (0, a), (– a, 0), (0, – a);

respectively as shown in figure then potential in minimum

at the point

a. A b. B

c. C d. D

14. A spherical capacitor consists of two concentric spherical

conductors of inner one of radius1R maintained at

potential1V and the outer one of radius

2R at potential 2.V

The potential centre ( )2 1R x R> > is:

a. ( )1 21

2 1

V Vx R

R R

−−

− b.

( ) ( )( )

1 1 2 2 2 1

2 1

V R R x V R x R

R R x

− + −

−

c. 21

1 1

V xV

R R+

− d. 1 2

2 1

V Vx

R R

++

15. Two identical short bar magnets, each having magnetic

moment M, are placed a distance of 2d apart with axes

perpendicular to each other in a horizontal plane. The

magnetic induction at a point midway between them is

a. 0

3( 2)

4

M

d

µπ

b. 0

3( 3)

4

M

d

µπ

c. 0

3

2 M

d

µπ

d. 0

3( 5)

4

M

d

µπ

16. The figure shows four wire loops, with edge lengths of

either L or 2L. All four loops will move through a region

of uniform magnetic field B

(directed out of the page) at

the same constant velocity. Rank the four loops according

to the maximum magnitude of the e.m.f. induced as they

move through the field, greatest first:

a. ( ) ( )c d a be e e e= < = b. ( ) ( )c d a be e e e= > =

c. c d b ae e e e> > >

d.

c d b ae e e e< < <

17. A resistance is connected to an AC source. The average

power dissipation in the resistance is

a. 50% of the peak power

b. 25% of the peak power

c. 41.4% of the peak power

d. equal to peak power

18. When a point source of monochromatic light is at a

distance of 0.2 m from a photoelectric cell, the cut-off

voltage and the saturation current are 0.6 V and 18 mA

respectively. If the same source is placed 0.6 m away from

the photoelectric cell, then

a. the stopping potential will be 0.2 V

b. the stopping potential will be 0.6 V

c. the saturation current will be 6 mA

d. the saturation current will be 18 mA

19. A beam of fast moving alpha particles were directed

towards a thin film of gold. The parts BA ′′, and C ′ of the

transmitted and reflected beams corresponding to the

incident parts A, B and C of the beam, are shown in the

adjoining diagram. The number of alpha particles in

a. B′ will be minimum and in C′ maximum

b. A′ will be maximum and in B′ minimum

c. A′ will be minimum and in B′ maximum

d. C′ will be minimum and in B′ maximum

B′ B A

C C′

A′

a b

c d

• • •

• • •

• • •

• • •

•

•

•

•

•

•

•

•

D

C

B E →

T1

T3

T2

I

λ

61 Mock Test-5

20. Energy released in the fission of a single 235

92 U nucleus

is 200 MeV. The fission rate of a 235

92 U fuelled reactor

operating at a power level of 5W is

a. 1101056.1 −+× s b. 1111056.1 −+× s

c. 1161056.1 −+× s d. 1171056.1 −+× s


21. A train of 150 m in length is going towards north direction

at a speed of 10 m/s. A parrot flies at a speed of 5 m/s

towards south direction parallel to the railway track. Then

the time taken by the parrot to cross the train is equal to:

a. 12 sec b. 8 sec

c. 10 sec d. 15 sec

22. What is the maximum value of the force F such that the

block shown in the arrangement, does not move?

a. 20 N b. 10 N

c. 12 N d. 15 N

23. When an astronaut in a rocket reaches near the moon, he

sends a radiowave of frequency 5000 MHz towards the

moon. He finds that the reflected wave from the moon has

a frequency 86 kHz more than the real frequency. The

velocity of the rocket with respect to the moon is:

a. 1.29 km/sec b. 2.58 km/sec

c. 3.87 km/sec d. 5.16 km/sec

24. A battery of 10 cells each of e.m.f. E = 1.5 V, and internal

resistance 0.5Ω has 1 cell wrongly connected. It is being

charged by 220V power supply with an external resistance

of 47 Ω in series. The potential difference across the

battery is:

a. 32 V b. 8 V

c. 180 V d. 188 V

25. A cell is connected between the points A and C of a

circular conductor ABCD of centre O with angle

A 60 .OC = ° If

1B and 2B are the magnitudes of the

magnetic fields at O due to the currents in ABC and ADC

respectively, the ratio 1

2

B

Bis

a. 0.2 b. 6

c. 1 d. 5

1A i2

300o B

C A

D

60o

i1

O

60°

F 3m kg=1

2 3µ =


Physics 62

JEE ADVANCE PAPER-I



1. Given,

1R 1= Ω

1C 2 Fµ= µF

2R 2= Ω

2C 4 F= µF

(I) (II) (III)

The time constants (in µS) for the circuits I, II, III are

respectively

a. 18, 8 / 9, 4 b. 18, 4, 8 / 9

c. 4, 8 / 9,18 d. 8 / 9,18, 4

2. A circuit is connected as shown in the figure with the

switch S open. When the switch is closed the total amount

of charge that flows from Y to X is

a. 0 b. 54 µC

c. 27 µC d. 81 µC

3. A parallel plate capacitor C with plates of unit area and

separation d is filled with a liquid of dielectric constant

2.K = The level of liquid is 3

dinitially. Suppose the

liquid level decreases at a constant speed ,V the time

constant as a function of time t is

a. 06

5 3

R

d Vt

ε+

b. 0

2 2 2

(15 9 )

2 3 9

d Vt R

d dVt V t

ε+− −

c. 06

5 3

R

d Vt

ε−

d. 0

2 2 2

(15 9 )

2 3 9

d Vt R

d dVt V t

ε−+ −

4. A 2µF

capacitor is charged as shown in figure. The

percentage of its stored energy dissipated after the switch

S is turned to position 2 is

a. 0% b. 20% c. 75% d. 80%

5. In the given circuit, a charge of +80 µF is given to the

upper plate of the 4µF

capacitor. Then in the steady state,

the charge on the upper plate of the 3µF capacitor is.

a. +32 µC b. +40 µC

c. +48 µC d. +80 µC

6. Two capacitors C1 and C2

are charged to 120 V and 200 V

respectively. It is found that by connecting them together

the potential on each one can be made zero. Then

a. 21 35 CC = b. 21 53 CC =

c. 053 21 =+ CC d. 21 49 CC =



7. The magnetic flux of φ (in weber) in a closed circuit of

resistance 3 ohm varies with time t (in second) according

to the equation 22 10 3.t tφ = − + What will be the

magnetic of induced current at t = 0.25 s ?

8. An a.c. generator gives an output voltage of E = 170 sin

56⋅52t. What is the frequency of alternating voltage

produced?

9. A condenser of capacitance 0.144 pH is used in a

transmitter to transmit at wavelength λ. If inductance of

2

1

πmH is used for resonance, what is the value of λ?

10. In the series RLC circuit as shown in Fig., what would be

the ammeter reading?

4 Fµ80 C+ µ

2 Fµ 3 Fµ

1 2

s

2µF 8µF V

R

C

d 3

d

3µF 6µF X

Y

9V

3Ω 6Ω

S

V

R2

R2

C1

C2 R1 R2 C1 C2

V

C1 C2

V R1

R2 + –

+ –

63 Mock Test-5

11. An a.c. source of frequency 1000 Hz is connected to a coil

of 200

πmH and negligible resistance. If effective current

through the coil is 7.5 mA, what is the voltage across the

coil?

12. Two alternating currents are given by ( )1 0 sinI I tω φ= −

and ( )2 0 cosI I tω φ= + . What is the ratio of virtual values

of the two currents?

13. A silver sphere of radius 1 cm and work function 4.7 eV is

suspended from an insulating thread in free-space. It is

under continuous illumination of 20 nm wavelength light.

As photoelectrons are emitted, the sphere gets charged

and acquires a potential. The maximum number of photo-

electrons emitted from the sphere is A 10ZA× (where 1 <

A < 10). The value of Z is

14. When a piece of metal is illuminated by monochromatic

light of wavelength λ then the stopping potential for

photoelectric current is 3V0. When the same surface is

illustrated by light of wavelength 2λ, then stopping

potential becomes the V0 value of Threshold wavelength

for photoelectric current is aλ, where the value of a is





A fixed thermally conducting cylinder

has a radius R and height 0.L The

cylinder is open at its bottom and has a

small hole at its top. A piston of mass

M is held at a distance L from the top

surface, as shown in the figure. The

atmospheric pressure is 0.P

15. The piston is now pulled out slowly and held at a distance

2L from the top. The pressure in the cylinder between its

top and the piston will then be

a. 0P b. 0

2

P

c. 0

22

P Mg

Rπ+ d. 0

22

P Mg

Rπ−

16. While the piston is at a distance 2L from the top, the hole

at the top is sealed. The piston is then released, to a

position where it can stay in equilibrium. In this condition,

the distance of the piston from the top is

a. 2

0

2

0

2(2 )

P RL

R P Mg

ππ

+ b.

2

0

2

0

(2 )P R Mg

LR P

ππ

−

c. 2

0

2

0

(2 )P R Mg

LR P

ππ

+

d. 2

0

2

0

(2 )P R

LR P Mg

ππ

−


The nuclear charge (Ze) is non-uniformly distributed within a

nucleus of radius R. The charge density ( )rρ [charge per unit

volume] is dependent only on the radial distance r from the

centre of the nucleus as shown in figure. The electric field is

only along the radial direction.

17. The electric field at r R= is

a. independent of a

b. directly proportional to a

c. directly proportional to 2a

d. inversely proportional to a

18. For 0,a = the value of d (maximum value of ρ as shown

in the figure) is

a. 3

3

4

Ze

Rπ b.

3

3Ze

Rπ

c. 3

4

3

Ze

Rπ d.

33

Ze

Rπ

o a R r

d

( )ρ r

200 V 200 V

100 V, 50 Hz

R = 500 Ω L C

~

A

L

L0

2

Piston


Physics 64




1. A parallel plate capacitor has a dielectric slab of dielectric

constant K between its plates that covers 1/3 of the area of

its plates, as shown in the figure. The total capacitance of

the capacitor is C while that of the portion with dielectric

in between is C1. When the capacitor is charged, the plate

area covered by the dielectric gets charge Q1 and the rest

of the area gets charge Q2. The electric field in the

dielectric is 1E and that in the other portion is E2.

Choose

the correct option/options, ignoring edge effects.

a. 1

2

1=E

E b. 1

2

1=

E

E K

c. 1

2

3=

Q

Q K d.

1

2 +=

C K

C K

2. A parallel plate capacitor having plates of area S and plate

separation d, has capacitance 1C in air. When two

dielectrics of different relative permittivities (ε1 = 2 and

ε2 = 4) are introduced between the two plates as shown in

the figure, the capacitance becomes2 .C The ratio 2

1

C

Cis

a. 6/5 b. 5/3 c. 7/5 d. 7/3

3. Consider a cylindrical element as shown in the figure.

Current flowing the through element is I and resistivity of

material of the cylinder is .ρ Choose the correct option out

the following

a. Power loss in first half is four times the power loss in

second half

b. Voltage drop in first half is twice of voltage drop in

second half

c. Current density in both halves are equal

d. Electric field in both halves is equal

4. Figure shows three resistor configurations R1, R2, and R3

connected to 3V battery. If the power dissipated by the

configuration R1, R2, and R3 is P1, P2, and P3, respectively,

then

a. 1 2 3P P P> > b. 1 3 2P P P> >

c. 2 1 3P P P> > d. 3 2 1P P P> >

5. Incandescent bulbs are designed by keeping in mind that

the resistance of their filament increases with the increase

in temperature. If at room temperature, 100 W, 60 W and

40 W bulbs have filament resistances R100, R60 and R40,

respectively, the relation between these resistances is

a. 100 40 60

1 1 1

R R R= + b.

100 40 60R R R= +

c. 100 60 40R R R> >

d.

100 60 40

1 1 1

R R R> >

6. To verify Ohm’s law, a student is provided with a test

resistor RT, a high resistance R1, a small resistance R2, two

identical galvanometers G1 and G2, and a variable voltage

source V. The correct circuit to carry out the experiment is

a. b.

c.

d.



7. Image of an object approaching a convex mirror of radius

of curvature 20 m along its optical axis is observed to

move from 25

3 m to

50

7 m in 30 seconds. What is the

speed of the object in km per hour?

G2

R2

RT

R1

V

G1

G2

R1

RT

R2

V

G1

G1

G2 R1

RT R2

V

G1

G2 R2

RT R1

V

R1 R2 R3

1

1

1

1

1

3V

1

1

1

1

1 3V 1Ω

1Ω

1Ω

1Ω

1Ω

3V

4r I

./ 2ℓ ./ 2ℓ

2r

A B C

d/2

d

S/2

S/2

ε1

ε2

2E–

+

–

+ –

+

65 Mock Test-5

8. Water (with refractive index = 4/3) in a tank is 18 cm

deep. Oil of refractive index 7

4lies on water making a

convex surface of radius of curvature ‘R = 6 cm’ as shown.

Consider oil to act as a thin lens. An object ‘S’ is placed

24 cm above water surface. The location of its image is at

‘x’ cm above the bottom of the tank. Then ‘x’ is

9. Consider a concave mirror and a convex lens (refractive

index = 1.5) of focal length 10 cm each, separated by a

distance of 50 cm in air (refractive index = 1) as shown in

the figure. An object is placed at a distance of 15 cm from

the mirror. Its erect image formed by this combination has

magnification 1.M When the set-up is kept in a medium of

refractive index 7/6, the magnification becomes 2.M The

magnitude 2

1

M

Mis

10. A monochromatic beam of light is incident at 60° on one

face of an equilateral prism of refractive index n and

emerges from the opposite face making an angle θ (n)

with the normal (see the figure). For 3=n the value of

θ is 60° and .d

mdn

θ= The value of m is

11. An α-particle and a proton are accelerated from rest by a

potential difference of 100 V. After this, their de Broglie

wavelengths are αλ and pλ respectively. The ratio ,

p

α

λ

λto

the nearest integer, is

12. A silver sphere of radius 1 cm and work function 4.7 eV is

suspended from an insulating thread in free space. It is

under continuous illumination of 200 nm wavelength

light. As photoelectrons are emitted, the sphere gets

charged and acquires a potential. The maximum number

of photoelectrons emitted from the sphere is A 10 z×

(where 1 < A < 10). The value of 'Z' is

13. A proton is fired from very far away towards a nucleus

with charge Q = 120 e, where e is the electronic charge. It

makes a closest approach of 10 fm to the nucleus. The de

Broglie wavelength (in units of fm) of the proton at its

start is: (take the proton mass,

27(5 / 3) 10 kg;pm −× 15/ 4.2 10h e −= × 9

0

1. / ; 9 10

4J s C

πε= ×

15/ ;1 10 )m F fm m

−=

14. The work functions of silver and sodium are 4.6 eV and

2.3 eV, respectively. The ratio of the slope of the stopping

potential versus frequency plot for silver to that of sodium

is



15. Some laws / processes are given in Column I. Match these

with the physical phenomena given in Column II.

Column I Column II

(A) Intensity of light

received by lens

1. Radius of aperture (R)

(B) Angular magnification 2. Dispersion of lens

(C) Length of telescope 3. Focal length ,a ef f

(D) Sharpness of image 4. Spherical aberration

a. A-1; B-3; C-3; D-1,2,3

b. A-1,2,3; B-4; C-2,3, D-1

c. A-1; B-3; C-1,2,3; D-2

d. A-2,3; B-1; C-4; D-3

16. Study of properties of matter helps the student to know

some phenomenon having important applications of the

principles learnt by him in his daily life. Can you match

some of the phenomenon you observe with the principles

you have studied.

Column I Column II

(A) A cricketer spins his ball

while bowling to change

the direction and

momentum of ball is

1. Surface tension

θ 60°

15 cm

50 cm

R 6cm=S

1.0µ =

7 / 4µ =

4 / 3µ =

Physics 66

according to principle of

(B) A painter askes you for

primers coating on walls

before panting to

remove sewage of water

through bricks in walls

2. Gravitational pull

(C) Formation of stars in

universe

3. Bernoulli’s pull

(D) Ships are asked to

spread oil on sea surface

during his tides

4. Capillarity

a. A-1; B-2; C-3; D-4 b. A-4; B-3; C-2; D-1

c. A-3; B-2; C-1; D-4 d. A-3; B-4; C-2; D-1

17. Match the statement of Column I with those in Column II:

Column I Column II

(A) Magnetic field intensity is

defined as

1. The spin motion

of electron

(B) Which of the following,

the most suitable material

for making permanent

magnet is

2. Steel

(C) In the case of bar magnet,

lines of magnetic

induction

3. Run continuously

through the bar

and outside

(D) A sensitive magnetic

instrument can be shielded

very effectively from

outside magnetic fields by

placing it inside a box of

4. Soft iron of high

permeability

a. A-4; B-3; C-2; D-1 b. A-2; B-4; C-3; D-1

c. A-1; B-2; C-3; D-4 d. A-4; B-1; C-3; D-2

18. Column I defines the type of modulus or coefficient of

elasticity. Column II gives the type of corresponding

modulus. Match the definition with proper type of

elasticity.

Column I Column II

(A) Ratio of longitudinal

or tensile stress to

longitudinal strain

1. Modulus of Rigidly

(B) Ratio of normal or

hydrostatic stress to

volumetric strain

2. Poisson’s ratio

(C) Ratio of lateral strain

to longitudinal strain

3. Bulk modulus

(D) Ratio of tangential

stress to shear strain

4. Young’s modulus

a. A-1; B-2; C-3; D-4 b. A-4; B-3; C-2; D-1

c. A-3; B-2; C-1; D-4 d. A-2; B-1; C-4; D-3


67 Mock Test-5

ANSWER & SOLUTIONS

JEE-Main

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

d a b a c b d b d c

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

b b a b d b a b b b

21. 22. 23. 24. 25.

c a d a c

1. (d) Velocity gradient 1

1[ ][ ]

[ ]

v LTT

x L

−−= = =

Potential gradient2 3 1

3 1[ ][ ]

[ ]

V ML T AMLT A

x L

− −− −= = =

Energy gradient 2 2

2[ ][ ]

[ ]

E ML TMLT

x L

−= = =

Pressure gradient1 2

2 2[ ][ ]

[ ]

P ML TML T

x L

− −− −= = =

2. (a) ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ4 3 2r a b c i j i j k i j k= + + = − − + − = + −

2 2 2

ˆˆ ˆˆ

| | 1 1 ( 1)

r i j kr

r

+ −= =

+ + −

ˆˆ ˆ

3

i j k+ −=

3. (b) Gravitational field is a conservative force field. In a

conservative force field work done is path independent

∴ 1 2 3

W W W= =

4. (a) The desired moment of inertia is

x l

x l

I dI

=+

=−

= ∫

( )2

2 2sin sin2 3

l

l

m mldx x

lα α

+

−

= = ∫

5. (c) 1 1,m M T T= =

⇒ 2 2

5,

3

Tm M m T= + =

⇒ 11 1

2 22

2 /

2 /

m KT m M

T m M mm K

π

π= = =

+

or,

2

2

1

5 /3 25

9

M m T T

M T T

+ = = =

⇒ 16

9

m

M=

6. (b) 4

0.410

g

g

Vf

r

∆= = =

∆ N/kg

7. (d) 3 34 4

3 3R n rπ π= ×

1/ 3( ) 10R n r r= =

Decrease in surface area 2 2(4 ) 4A n r Rπ π∆ = −

2 21000(4 ) 4 (10 )r rπ π= −

2 24 [1000 100] 900 4r rπ π= − = ×

Energy released A T= ∆ × =2900 4 T

calr

J

×=

π

Let θ be the rise in temperature;’ then

2

34 900 4 T1000 1

3

rr d

J

×× × ×

ππ θ =

∴ 2.7T

dJr

=θ

8. (b) For simple pendulum 2l

Tg

π= when weight

suspended on elastic wire is increased, it length increases,

so time period increases.

9. (d) Zero Kelvin –237 C= ° (absolute temperature). As no

matter can attain this temperature, hence temperature can

never be negative on Kelvin scale.

10. (c) For an ideal gas : pV = nRT

For p = constant p V nT T∆ = ∆

∴ V nR nR V

nRTT p T

V

∆= = =

∆

∴ 1V

V T T

∆=

∆ or

1

Tδ =

Therefore, δ is inversely proportional to temperature T.

i.e., when T increases, δ decreases and vice-versa.

Hence, Tδ − graph will be a rectangular hyperbola as

shown in the above figure.

11. (b) 21 1

2 2mv J ms⋅ = ∆

∆θ

⇒ 2

4

v

Js=∆θ

12. (b) Wien’s displacement law for a perfectly black body is

mTλ = constant = Wien’s constant b

Here, mλ is the minimum wavelength corresponding to

maximum intensity I.

• x

dx

α x sin α

T

Physics 68

or 1

mT

λ ∝

From the figure ( ) ( ) ( )1 3 2m m mλ λ λ< <

Therefore, 1 3 2T T T> >

13. (a) In the direction of electric field, potential decreases.

14. (b) Let 1q and 2q be charges on inner and outer spherical

conductors.

Then 1 21

0 1 2

1

4

q qV

R Rπε

= +

. . .(i)

Also 1 22

0 2

1

4

q qV

Rπε+

= . . .(ii)

Then potential at point P, 1 2

0 2

1

4

q qV

x Rπε

= +

. . .(iii)

Solving (i) and (ii), 1q and 2q may be found. If these are

substituted in (iii), we get

( ) ( )

( )1 1 2 2 2 1

2 1

V R R x V R x RV

R R x

− + −=

−

15. (d) At point P net magnetic field 2 2

1 2netB B B= +

where 01 3

2.

4

MB

d

µπ

= and 02 3

.4

MB

d

µπ

=

⇒

0

3

5.

4net

MB

d

µπ

=

16. (b) Emf induces across the length of the wire which cuts

the magnetic field.

(Length of c = Length d) > (Length of a = b).

So, ( )c de e= ( )a be e> =

17. (a) Peak power, 0 0 0P V I=

Average power dissipation, 0 0 cos2

av

V IP φ=

For a resistive circuit 0φ =

⇒ cos 1φ =

⇒ 0 0 0 50%2 2

av

V I PP = = = of

0P

18. (b) By changing distance the intensity change but

frequency remains same; so stopping potential remains

same.

19. (b) Because atom is hollow and whole mass of atom is

concentrated in a small centre called nucleus.

20. (b) 200Fission

Energy=

200 MeV

196106.110200

−×××= J

Fission rate 5

200 MeV=

111056.1 ×= fission/sec.

21. (c) Relative velocity of parrot and train

10 5 15 / = + = m/s

∴ Time 150

1015

s

v= = = 10 s

22. (a) Free body diagram (FBD) of the block (shown by a

dot) is shown in figure. For vertical equilibrium of the

block

sin 60N mg F= + ° 3 32

Fg= +

For no motion, force of friction or cos 60f F≥ °

⇒ cos 60N F≥ °µ or 1 3

32 22 3

F Fg

+ ≥

or 2

Fg ≥

or 2 or 20NF g≤

Therefore, maximum value of F is 20 N.

23. (d) Here change in frequency takes place in two steps

∴ 2v

v vc

∆ =

⇒ 3 6

86 10 2 5000 10v

c× = × ×

⇒

3

10

86 10

10v

×=

7 886 10 3 10c −= × × × m/s

3

2.58 10= × m/s

24. (a) Current in circuit

( )220 9 1 1 5

47 0 5 10i

− − × ⋅=

+ ⋅ ×

220 124 A

52

−= =

∴ Potential difference across battery which is being charged

V E ir= +

( ) ( )8 1 5 4 0 5 10 32= × ⋅ + × ⋅ × = V

Vertical

Horizontal

mg + F sin 60°

N

f F cos 60°

P

B2

B1

d

S

N

1

2

N

S

d

69 Mock Test-5

25. (c) 0

4

iB B i

r= ⇒

µ θ∝ θ

π (but 1 2 2

2 1 1

i l

i l= =

1

2θ

θ)

⇒

1 1

2 2 2

.B i

B i=

1θ

θ

So, 1

2 2 1

B

B= ×

1 2θ θ

θ θ

⇒1 2B B=

JEE Advance Paper-I

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

d c a d c b 3 9 7 2

11. 12. 13. 14. 15. 16. 17. 18.

3 1 7 4 a d a b

1. (d) 1 8 / 9τ µ= µ S

⇒ 2 18τ µ= µ S ⇒

3 4τ µ= µ S

2. (c) 27 Cµ

Initial charge distribution (when switch S is open)

Final charge distribution (when switch S is closed)

3. (a)

0 0

equivalent0 0

2.2

3 32

2

3 3

d dvt vt

C

d dvt vt

ε ε

ε ε

− +=

+− +

∴ equivalentC Rτ =

4. (d) 21(2)

2iU V= ×

⇒ 21 (2) (8)

2 10lossU V

×∆ = × ×

⇒ 4

80%5

losslost

i

U

Uη

∆= = =

5. (c) Charge on capacitor of 3 Fµ

⇒ 3

80 48 C2 3

Q = = µ+

6. (b) Potential 0= on connecting them together i.e., 0=Q

i.e. 2211 VCVC = [capacitance is positive but they are

connected with opposite polarity]

⇒ 21 200120 CC =

⇒ 21 53 CC =

7. (3) Here, 3R = Ω

⇒⇒⇒⇒ 22 10 3t tφ = − +

? 0.25I t= = g

⇒⇒⇒⇒ 2(2 10 3)d d

e t tdt dt

φ= = − − +

⇒⇒⇒⇒ e = – 4 t + 10

At 0.25 , 4 0.25 10 9 voltt s e= = − × + =

⇒⇒⇒⇒ 9

3 A3

eI

R= = =

8. (9) Here, 170sin 56.52E t= Compare with the standard form of equation of alternating

emf

0 sinE E tω=

⇒ 56.52ω =

⇒ 2 56.52vπ =

⇒ 56.52 56.52

92 2 3.14

vπ

= = =×

Hz.

9. (7) Here, 120.144pF 0.144 10C F−= = × F

⇒ 3

2 2

1 10mH HL

π π

−

= =

⇒ 1

2 LCυ

π=

⇒ .2v

LCλ υ πυ

= =

3

8 12

2

103 10 2 0.144 10π

π

−−= × × × ×

⇒ 8

8 1.2 103 10 2λ π

π

−×= × × ×

3µF 6µF X

Y

9V 3Ω 6Ω

S

+9µC +18µC

+27µC

+9µC +36µC

3µF 6µF X

Y

9V

3Ω 6Ω

S

+18µC +18µC

1A

1A

i2

300o 1

2

60o

i1

O

+ –

Physics70

The correct answer is 7 m.

10. (2) As

∴ The circuit is series resonance circuit Z = R

⇒

11. (3) Here, v = 1000 Hz,

As resistance of coil is negligible,

∴

12. (1)

Phase difference has no impact on virtual value of current.

13. (7) Here, eV,

Let be the stopping potential. According to Einstein’s

photoelectric equation (using hc = 1240 eV nm), we have

1240eV

eV 4.7eV200nm

s

hc− = −

0φ

λ

6.2eV 4.7eV 1.5eV 1.5eV= − = =

∴

The sphere will stop emitting photoelectrons, when the

potential on its surface becomes 1.5 V. Let n be the no. of

photoelectrons emitted from sphere.

The charge on sphere,

∴

As per questions,

14. (4) 03eV

and 0eV

or 02eV or 0eV

hc

If is the stopping wavelength, then 0eV

or

15. (a) Po

16. (d) 2 2

0( )Mg P R P Rπ π+ =

⇒ 2 2

0(2 ) ( )P L R P x Rπ π=

1 1 2 2(PV PV= for isothermal process)

⇒ 2

0

2

0

(2 )P R

x LR P Mg

ππ

= −

17. (a) independent of a

18. (b) 2

0

( )42

Rd

q R x x dx Zeπ= − =∫

⇒ 3

3Zed

Rπ=

JEE Advance Paper-II

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

a,d d b c d c 3 2 7 2

11. 12. 13. 14. 15. 16. 17. 18.

3 7 7 1 a d c b

1. (a, d) As E = V/d

1 2/ 1=E E (both parts have common potential difference)

Assume 0C be the capacitance without dielectric for

whole capacitor.

⇒ 0 02

3 3+ =

C Ck C

⇒ 1

2++

C k

C k⇒ 1

2

.2

+Q k

Q

2. (d)0

010

442

/ 2

ε ε= =

SS

Cd d

⇒ 0 020 30

2,

ε ε=

S SC C

d d

⇒ '

10 1010

1 1 1

C CC= +

0

11

2 2

d

Sε = +

⇒ 10

' 04

3

ε=

SC

d

⇒ ' 02 30 10

7

3

ε= + =

SC C C

d

⇒ 2

1

7

3=

C

C

7.2 mλ =

200VL CV V= =

1002A

50

vE E

IZ R

υυ = = = =

200 0.2mH = HL

π π=

37.5mA 7.5 10 AIυ−= = ×

?Eυ =

( ) ( )2L

E I X I vLυ υ υ π= =

3 0.27.5 10 2 1000 3Vπ

π− = × × × =

1

2

0

0

/ 21

/ 2

I I

I I

υ

υ

= =

2

01 cm 10 m; 4.7 ,r eVφ−= = = 200nm.λ =

sV

1.5sV =

, 1.5Vsq ne V= =

270

19 9

4 (1.5) (10 ) 11.05 10

1.6 10 (9 10 )

sV rn

e

πε −

−

× × ×= = × = ×

× ×710 1.05 10 or 7ZA Z× = × =

0 03hc

eV φλ

= − 0 02

hceV φ

λ= −

0

12 1

2 2

hc hfeV

λ λ = − = 0

4hc

eVλ=

0λ 0 0/eV hc λ=

0

0

4hc

eVλ λ= =

10C 20C

30C

71 Mock Test-5

3. (b) 1 1

2 2

4

1= =

R A

R A

⇒ 2

1 1

2

2 2

4

1= =

P I R

P I R

⇒ 1 1

2 2

4

1= =

V IR

V IR⇒ 1

2

1

4=

J

J

4. (c)2V

PR

=

⇒ 1 21 ,R R= Ω 31/ 2 , 2R= Ω = Ω

∴ 2 1 3P P P> >

5. (d) Power 1/ R∝

6. (c) 1G is acting as voltmeter and 2G is acting as ammeter.

7. (3) For 1

50,

7v m= 1 25mu = −

⇒ 2 2

25, 50m

3v m u= = −

Speed of object 25 18

330 5

= × = kmph.

8. (2)1

71

7 1 4

4 24 6V

−− =

−

⇒ 1 21 cmV =

⇒ 2

4 / 3 7 / 40

21V− =

⇒ 2 16 cmV =

⇒ 18 16 2 cmx = − =

9. (7) Image by mirror is formed at 30 cm from mirror at its

right and finally by the combination it is formed at 20 cm

on right of the lens. So in air medium, magnification by

lens is unity. In second medium, 7

,6

µ = focal length of

the lens is given by,

1 2

1 2

1 11 (1.5 1)101 1.5 1 1

17 / 6

R R

f R R

− −

= − −

⇒ 35

2f cm= cm

So, in second medium, final image is formed at 140 cm to the

right of the lens. Second medium does not change the

magnification by mirror.

So 2 2

1 1

2

1

7m

m

M MM

M M M= =ℓ

ℓ

10. (2) Snell’s Law on 1st surface: 1

3sin

2= n r

1

3sin

2=r

n . . .(i)

⇒

2

1 2

3 4 3cos 1

4 2

−= − =

nr

n n

1 2 60+ = °r r . . .(ii)

Snell’s Law on 2nd

surface: 2sin sinn r = θ

Using equation (i) and (ii) 1sin(60 ) sinn r° − = θ

1 1

3 1cos sin sin

2 2n r r

− = θ

( )234 3 1 cos

4

d dn

dn dn

θ− − = θ

For 60θ = ° and 3=n

⇒ 2d

dn

θ=

11. (3) 21

2mv qV=

⇒ h

mvλ =

⇒ 8 3.= ≃λ

12. (7) Stopping potential Whc

λ= −

6.2 eV 4.7 eV= − 1.5 eV=

⇒ 1.5Kq

Vr

= =

⇒ 2

7

9 19

1.5 101.05 10

9 10 1.6 10n

−

−

×= = ×

× × ×

⇒ 7Z =

13. (7) 2

2

0

1 1 (120)

2 4 (10 )

emv

fmπε∞ =

[By Conservation of energy]

Assuming the nucleus to be considerably massive, we can

disregard its motion.

∴ Let momentum of proton be p mv∞=

∴ 2 2

15

0

1 (120)

2 4 (10 10 )p

p e

m πε −=

×

∴ 2

9 27

14

5 1209 10 2 10

3 10

ep −

−

×= × × × × ×

Physics 72

⇒ 9 27 14 230 120 10p e− += × × ×

∴ 4 23600 10p e−= × ×

∴ 260 10p e

−= × ×

∴ h

pλ =

560 10

h

e=

× ×

1515

2

42 10 4210 m

660 10

−−

−

×= = ×

×

∴ 7fm=λ

14. (1) Slope of graph is h/e = constant ⇒ 1

15. (a) A→1, B→3, C→3, D→1,2,3

A, C and D in case of concave mirror convex lens image

can be real, virtual, diminished magnified or of same size.

B is case of convex mirror image is always virtual (for

real object)

16. (d) A-3; B-4; C-2; D-1

According to Bernoulli’s theorem the spinning of ball

along with horizontal throw will change the velocity of

streamlines of air above and below the ball which will

change the pressure on ball above and below it which

cause change in its direction of motion. This is called

magnus effect.

Primers contain materials which make obtuse angle of

contact due to their surface tension and do not allow

capillary rise of water in pours of bricks.

Interstellar dust particles attract each other by

gravitational pull to build a star.

Oil has smaller surface tension than water and is lighter

than water so it spreads move over the surface of water to

decrease height of tides.

17. (c) A-1; B-2; C-3; D-4

18. (b) A-4; B-3; C-2; D-1

Longitudinal stress

Longitudinal strain

F lY

A l= × = =

∆Young’s modulus

Normal stress

Volumetric strain

F VK

A V= × = =

∆ Bulk modulus

= Volume elasticity

Poisson’s ratio Lateral strain

Longitudinal strain=

⇒

DD lD

l D l

l

σ

∆∆

= = ×∆ ∆

73Mock Test-1

Mock Test 1 “JEE-Main”





prohibited.






questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response.

6. Candidates will be awarded marks as stated above in instruction No. 5 for correct response of each question. 1/4 (one

fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will

be made if no response is indicated for an item in the answer sheet.

7. There is only one correct response for each question. Filling up more than one response in each question will be treated as

wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.

8. Use Blue/Black Ball Point Pen only for writing particulars/ marking responses on Side-1 and Side-2 of the Answer Sheet.






11. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room / Hall.







Roll Number : in figures

in words




A Test Booklet code

Chemistry74

Read the following instructions carefully:







5. For each incorrect response, one-fourth (¼) of the total marks allotted to the question would be deducted from

the total score. No deduction from the total score, however, will be made if no response is indicated for an item

in the Answer Sheet.











Invigilator on duty and sign the Attendance Sheet again. Cases where a candidate has not signed the Attendance

Sheet a second time will be deemed not to have handed over the Answer Sheet and dealt with as an unfair

means case. The candidates are also required to put their left hand THUMB impression in the space

provided in the Attendance Sheet.

12. Use of Electronic/Manual Calculator and any Electronic Item like mobile phone, pager etc. is prohibited






hall/room.

75Mock Test-1

JEE-MAIN: CHEMISTRY MOCK TEST-1

1. In a galvanic cell, the salt bridge a. does not participate chemically in the cell reaction. b. stops the diffusion of ions from one electrode to another. c. is necessary for the occurrence of the cell reaction d. ensures mixing of the two electrolytic solutions

2. Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately

a. 1.86 Å b. 3.22 Å c. 5.72 Å d. 0.93 Å

3. Which of the following is the energy of a possible excited state of hydrogen?

a. 13.6 eV b. 6.8 eV c. 3.4 eV d. 6.8 eV

4. The number of atoms in 100 g of an fcc crystal with

density –3=10.0 g cm and cell edge equal to 200 pm is

equal to a. 245 10 b. 255 10 c. 236 10 d. 252 10

5. The following reaction is performed at 298 K.

2 22NO(g) O (g) 2NO (g)

The standard free energy of formation of NO (g) is 86.6

kJ/mol at 298 K. What is the standard free energy of

formation of 2NO (g) at 298 K? 12p(K 1.6 10 )

a. 12R(298) n (1.6 10 ) 86600 b. 1286600 R(298) n (1.6 10 )

c. 12n (1.6 10 )

86600R(298)

d. 120.5[2 86600 R(298) n (1.6 10 )]

6. The vapour pressure of acetone at 20 C is 185 torr. When

1.2 g of a non-volatile substance was dissolved in 100 g of

acetone at 20 C, its vapour pressure was 183 torr. The

molar mass (g mol–1) of the substance is:

a. 32 b. 64 c. 128 d. 488

7. The standard Gibbs energy change at 300 K for the

reaction 2A B C is 2494.2 J. At a given time, the

composition of the reaction mixture is 12[A] , [B] 2 and

12[C] . The reaction proceeds in the:

[ R 8.314 J / K / mol, e 2718 ]

a. forward direction because cQ K

b. reverse direction because cQ K

c. forward direction because cQ K d. reverse direction because cQ K

8. Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is:

a. 0 g b. 63.5 g c. 2 g d. 127 g

9. What is the half life period of a radioactive substance if 87.5% of any given amount of the substance disintegrates in 40 minutes?

a. 160 min b. 10 min c. 20 min d. 13 min 20 sec

10. 3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06 N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is :

a. 18 mg b. 36 mg c. 42 mg d. 54 mg

11. The ionic radii (in Å) of 3 2N , O and F are respectively:

a. 1.36, 1.40 and 1.71 b. 1.36, 1.71 and 1.40 c. 1.71, 1.40 and 1.36 d. 1.71, 1.36 and 1.40

12. In the context of the Hall-Heroult process for the extraction of A , which of the following statements is

false?

a. CO and 2CO are produced in this process

b. 2 3A O is mixed with 2CaF which lowers the melting

point of the mixture and brings conductivity

c. 3A is reduced at the cathode to form A

d. 3 6Na A F serves as the electrolyte

13. From the following statements regarding 2 2H O , choose

the incorrect statement a. It can act only as an oxidising agent b. It decomposes on exposure to light c. It has to be stored in plastic or wax lined glass bottles in

dark d. It has to be kept away from dust 14. Which one of the following alkaline earth metal sulphates

has its hydration enthalpy greater than its lattice enthalpy?

a. 4CaSO b. 4BeSO

c. 4BaSO d. 4SrSO

Chemistry76

15. Which among the following is the most reactive?

a. 2Cl b. 2Br c. 2I d. ICl

16. Which one has the highest boiling point? a. He b. Ne c. Kr d. Xe

17. The number of geometric isomers that can exist for square

planar [Pt(Cl)(py) 3 2(NH )(NH OH)] is (py = pyridine):

a. 2 b. 3 c. 4 d. 6

18. The color of 4KMnO is due to

a. M L charge transfer transition

b. d d transition

c. L M charge transfer transition

d. * transition

19. In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is: (Atomic mass

Ag 108, Br 80)

a. 24 b. 36 c. 48 d. 60

20. In the reaction shown below, the major product(s) formed is/are

a.

b.

c.

d.

21. Which compound would give 5-keto-2-methyl hexanal

upon ozonolysis?

a.

b.

c.

d.

22. The synthesis of alkyl fluorides is best accomplished by :

a. Free radical fluorination

b. Sandmeyer’s reaction

c. Finkelstein reaction

d. Swarts reaction

23. In the following sequence of reactions:

Toluene 2KMnO 2 2

4

SOCl H / Pd

BaSOA B C, the

product C is:

a. 6 5C H COOH b. 6 5 3C H CH

c. 6 5 2C H CH OH d. 6 5C H CHO

24. In the reaction

2NaNO / HCl

0 5 C CuCN/ KCN2D E N The product E

is

a.

b.

c.

d.

25. Which polymer is used in the manufacture of paints and

lacquers?

a. Bakelite b. Glyptal

c. Polypropene d. Poly vinyl chloride

CHH3

CN

CH3

CH3

COOH

CH3

NH3

CH3

3CH

3H C

3CH

3CH

3CH

3CH

3CH

3CH

3 3N H CH COO

3CH

O O

H|N 2+ H O

H|N

3CH

O

3CH

O O

H|N 2+ H O

3CH

O O

H|N

2NH

H|N

O

2NH

O

3CH

3+ CH COOH

77Mock Test-1

26. Which of the vitamins given below is water soluble? a. Vitamin C b. Vitamin D c. Vitamin E d. Vitamin K 27. Which of the following compounds is not colored yellow? a. 2 6Zn [Fe(CN) ] b. 3 2 6K [Co(NO ) ]

c. 4 3 3 10 4(NH ) [As(Mo O ) ] d. 4BaCrO 28. Match the statement of Column with those in Column II:

Column I Column II

(A)

1. Pseudo first order

(B)

2. Zero order

(C) 3. Second order

(D) 4. First order

a. A1; B3, 4; C2; D3 b. A2; B4; C3; D1 c. A1; B3, 2; C2; D4 d. A4; B1; C3; D2

Assertion and Reason

Note: Read the Assertion (A) and Reason (R) carefully to mark the correct option out of the options given below: a. If both assertion and reason are true and the reason is the

correct explanation of the assertion. b. If both assertion and reason are true but reason is not the

correct explanation of the assertion. c. If assertion is true but reason is false.

d. If the assertion and reason both are false.

e. If assertion is false but reason is true

29. Assertion: Enthalpy and entropy of any elementary

substance in the standard state are taken as zero.

Reason: At zero degree absolute, the constituent particles

become completely motionless.

30. Assertion: Molecularity has no meaning for a complex

reaction.

Reason: The overall molecularity of a complex reaction is

equal to the molecularity of the slowest step.

H12 22 11 2C H O H O

6 12 6C H O 6 12 6C H OHOH

3 2 5 H or OHCH COOC H

3 2 5CH COOH C H OHhv

2 2H Cl 2HCl

3 3CH Cl OH CH OH Cl


Chemistry78

JEE ADVANCE PAPER-I

Time 3 Hours. Max. Marks 264 (88 for Chemistry)

Read The Instructions Carefully



2. Section 1 contains 8 questions. The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).

Marking Scheme: +4 for correct answer and 0 in all other cases.


Marking Scheme: +4 for correct answer, 0 if not attempted and –2 in all other cases.

4. Section 3 contains 2 “match the following” type questions and you will have to match entries in Column I with the entries

in Column II.

Marking Scheme: for each entry in Column I, +2 for correct answer, 0 if not attempted and – 1 in all other cases.

NOTE: It’s the mock test as per previous year’s papers but sometimes IIT changes the test paper pattern and

marking scheme too.


This section contains EIGHT questions.

The answer to each question is a SINGLE DIGIT INTEGER

ranging from 0 to 9, both inclusive.

For each question, darken the bubble corresponding to the correct

integer in the ORS.

Marking scheme:

+4 If the bubble corresponding to the answer is darkened.

0 In all other cases.

1. If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride-ammonia complex (which behaves as

a strong electrolyte) is 0.0558 C, the number of

chloride(s) in the coordination sphere of the complex is

[ fK of water = 1.86 K kg mol–1]

2. The total number of stereoisomers that can exist for M is

3. The number of resonance structures for N is

4. The total number of lone pairs of electrons in 2 3N O is

5. For the octahedral complexes of 3Fe in SCN

(thiocyanato-S) and in CN ligand environments, the

difference between the spin-only magnetic moments in

Bohr magnetons (When approximated to the nearest

integer) is [Atomic number of Fe = 26]

6. Among the triatomic molecules/ions, 2 3 2BeCl , N , N O,

2 3 2 2 3NO ,O ,SCl , ICl , I and 2XeF , the total number of

linear molecule(s)/ion(s) where the hybridization of the

central atom does not have contribution from the d-

orbital(s) is

[Atomic number: S = 16, Cl = 17, I = 53 and Xe = 54]

7. Not considering the electronic spin, the degeneracy of the

second excited state (n = 3) of H atom is 9, while the

degeneracy of the second excited state of H is

8. All the energy released from the reaction X Y. 0rG

1193 kJ mol is used for oxidizing M as 3M M 02e ,E 0.25V. Under standard conditions, the

number of moles of M oxidized when one mole of X is

converted to Y is –1[F 96500 C mol ]

OH NaOH N

3CH

OM

3H C

3H C

79Mock Test-1


This section contains TEN questions.

Each question has FOUR options (a), (b), (c) and (d). ONE OR

MORE THAN ONE of these four option(s) is (are) correct.

For each question, darken the bubble(s) corresponding to all the

correct option(s) in the ORS.

Marking scheme:

+4 If only the bubble(s) corresponding to all the correct

option(s) is(are) darkened.

0 If none of the bubbles is darkened

–2 In all other cases

9. For the reaction: 3 2 4I ClO H SO 4 2HSO I

The correct statement(s) in the balanced equation is/are

a. Stoichiometric coefficient of 4HSO is 6

b. Iodide is oxidised c. Sulphur is reduced

d. 2H O is one of the products.

10. Compound(s) that on hydrogenation produce(s) optically inactive compound(s) is (are)

a. b.

c. d.

11. The major product of the following reaction is

a.

b.

c.

d.

12. In the following reaction, the major product is

a.

b.

c.

d.

13. The correct statement(s) for orthoboric acid is/are a. It behaves as a weak acid in water due to self ionisation b. Acidity of its aqueous solution increases upon addition

of ethylene glycol c. It has a three dimensional structure due to hydrogen

bonding. d. It is a weak electrolyte in water

14. The major product of the reaction is

a.

b.

c.

d.

15. The reactivity of compound Z with different halogens under appropriate conditions is given below:

The observed pattern of electrophilic substitution can be

explained by a. the steric effect of the halogen b. the steric effect of the tert-butyl group c. the electronic effect of the phenolic group d. the electronic effect of the tert-butyl group

16. An ideal gas in thermally insulated vessel at internal

pressure 1P , volume 1V and absolute temperature

1T expands irreversibly against zero external presssure,

as shown in the diagram. The final internal pressure,

volume and absolute temperature of the gas are 2 2P , V

and 2T , respectively. For this expansion.

extP 0

1 2 2P ,V ,T

extP 0

1 2 2P ,V ,T

Thermal insulation

Z 3 3C(CH )

OH2X

2 2mono halo substituted derivative when X I

2 2di halo substituted derivative when X Br

2 2tri halo substituted derivative when X Cl

3H C

3CH

2NH

OH

3H C

3CH

2CO H

OH

3H C

3CH

2CO H

OH

3H C 2NH

3CH OH

3H C

3CH

2CO H

2NH

2NaNO , aqueous HCl0 C

Br3H C

3CH

Br3H C

3CH

Br3H C

3CH3CH

Br3H C

3CH

2CH

3H C

3CH1 equivalent HBr

O3CH

O

3CH

O3CH

O 3CH

O

O3CH 2i. KOH , H O

ii. H , heat

HBr 3CH

3H C

Br H

3CH3H C

3CH

BrH 3CH2H C

Br H

3CH3H C

Chemistry80

a. q 0 b. 2 1T T c. 2 2 1 1P V P V d. 2 2 1 1P V P V

17. 3Fe is reduced to 2Fe by using

a. 2 2H O in presence of NaOH

b. 2 2Na O in water

c. 2 2H O in presence of 2 4H SO d. 2 2Na O in presence of 2 4H SO

18. The % yield of ammonia as a function of time in the

reaction 2 2 3N (g) 3H (g) 2NH (g), H 0 at 1(P,T )

is given below:

If this reaction is conducted at 2(P,T ), with 2 1T T , the

% yield of ammonia as a function of time is represented by

a. b.

c. d.


This section contains TWO questions.

Each question contains two columns, Column I and Column II

Column I has four entries (A), (B), (C) and (D)

Column II has five entries (1), (2), (3), (4) and (5)

Match the entries in Column I with the entries in Column II

One or more entries in Column I may match with one or more

entries in Column II

The ORS contains a 4 5 matrix whose layout will be similar to

the one shown below:

(A) (1) (2) (3) (4) (5)

(B) (1) (2) (3) (4) (5)

(C) (1) (2) (3) (4) (5)

(D) (1) (2) (3) (4) (5)

For each entry in Column I, darken the bubbles of all the

matching entries. For example, if entry (A) in Column I, matches

with entries (2), (3) and (4), then darken these three bubbles in the

ORS. Similarly, for entries (B), (C) and (D).

Marking scheme:

For each entry in Column I

+2 If only the bubble(s) corresponding to all the correct match(es)

is(are) darkened



19. Different possible thermal decomposition pathways for peroxyesters are shown below. Match each pathway from Column -I with an appropriate structure from Column –II is:

Column I Column II

(A) Pathway A 1.

(B) Pathway B 2.

(C) Pathway C 3.

(D) Pathway D 4.

a. →1; B→3; C→4; D→2

b. →2; B→4; C→4; D→1

c. →1; B→1; C→2; D→4

d. →3.; B→2; C→1; D→3

20. Match the orbital overlap figures shown in Column -I with the description given in Column -II and select the correct

OO

RR

O

(Peroxyester)

1

PCO

R R O

1

QCO

R R O

1 1

R2CO CO

RCO R O

S2RCO R O

R X carbonyl compound

R X carbonyl compound

1COR R O

% Yield

time

1T

2T

% Yield

time

1T

2T

% Yield

time

1T

2T

% Yield1T

time

1T

time

3CH

O

6 5 2C H CH

OO

3CH

O

6 5C H

OO

3CH

O

6 5 2C H CH

OO

3CH

2 6 5CH C H

3CH

O

6 5C HO

O3CH

6 5C H

81Mock Test-1

answer using the code given below the lists.

2 2 2 2(en H NCH CH NH ; atomic numbers :

Ti 22, Cr 24, Co 27; Pt 78)

Column I Column II

(A)

1. p-d antibonding

(B)

2. d-d bonding

(C)

3. p-d bonding

(D)

4. d-d antibonding

a. →2; B→3; C→1; D→4

b. →2; B→4; C→4; D→1

c. →1; B→1; C→2; D→4

d. →3.; B→2; C→1; D→3


Chemistry82


Time 3 Hours. Max. Marks 240 (80 for Chemistry)








4. Section 3 contains 2 “paragraph” type questions. Each paragraph describes an experiment, a situation or a problem. Two

multiple choice questions will be asked based on this paragraph. One or more than one option can be correct.

Marking Scheme: +4 for correct answer, 0 if not attempted and – 2 in all other cases.


marking scheme too.






integer in the ORS.

Marking scheme:



1. In dilute aqueous 2 4H SO , the complex diaquodioxalato-

ferrate (II) is oxidized by 4

MnO .− For this reaction, the

ratio of the rate of change of [H ]+ to the rate of change of

4[MnO ]− is

2. The number of hydroxyl group(s) in Q is

3. Among the following, the number of reaction(s) that

produce(s) benzaldehyde is

I

II

III

IV

4. The crystal of a solid is square packing of identical

spheres in each layer and spheres of one layer are placed

just above the voids made by spheres in previous layer.

The packing efficiency of such crystal (in %) is

[ 3.15, 2 1.4]π = =

5. Among the complex ions, 2 2 2[Co(NH - CH - CH -NH2)2

2Cl ] ,+ 3–

2 2 4 2[CrCl (C O ) ] , 2 4 2[Fe(H O) (OH) ] ,+ [Fe(NH3)2]

4(CN) ] ,−

2 2 2 2 2[Co(NH CH CH NH )− − −2

3(NH )Cl]+ and

2

3 4 2[Co(NH ) (H O)Cl] ,+ the number of complex ion(s) that

show(s) cis-trans isomerism is

2CO Me

2

DIBAL-H

Toluene, 78 C H O− °→

COCl2

4

H

Pd BaSO−→

2CHCl2H O

100 C°→

3

CO, HCl

Anhydrous AlCl / CuCl→

4aqueousdiluteKMnO (excess)H

heat 0 CP Q

+

°→ →

3H C

H

HO

3CH

83Mock Test-1

6. Three moles of 2 6B H are completely reacted with

methanol. The number of moles of boron containing

product formed is

7. The molar conductivity of a solution of a weak acid HX

(0.01 M) is 10 times smaller than the molar conductivity

of a solution of a weak acid HY (0.10 M). If 0 0

X Y,

the difference in their apK values, apK (HX) apK (HY),

is (consider degree of ionisation of both acids to be << 1)

8. A closed vessel with rigid walls contains 1 mol of 23892 U

and 1 mol of air at 298 K. Considering complete decay of 238

92 U to 20682 Pb, the ratio of the final pressure to the initial

pressure of the system at 298 K is







Marking scheme:

+4 If only the bubble(s) corresponding to all the correct option(s)

is(are) darkened.



9. Which of the following compounds will exhibit

geometrical isomerism?

a. 1-Phenyl-2-butene

b. 3-Phenyl-1-butene

c. 2-Phenyl-1-butene

d. 1,1-Diphenyl-2-propane

10. In the following reactions, the product S is

a.

b.

c. d.

11. The major product U in the following reactions is

a.

b.

c.

d.

12. In the following reactions, the major product W is

a.

b.

c.

d.

13. The correct statement(s) regarding, (i) HClO, (ii) 2HClO ,

(iii) 3HClO and (iv) 4HClO , is (are)

a. The number of Cl = O bonds in (ii) and (iii) together is two

b. The number of lone pairs of electrons on Cl in (ii) and (iii) together is three

c. The hybridisation of Cl in (iv) is 3sp

d. Amongst (i) to (iv), the strongest acid is (i)

N N

OH

N NOH

N N

OH

N N OH

2NHOH

NaOH 2NaNO , HCl

0 CV W

OH

HCH2

CH2

O

H|

O

O

H3C CH3 O

HCH3

O

H|

O

2 3 2CH CH CH , H radical initiator, O

high pressure, heat T U

3H C

NN

3H C

3H C NN 3H C

3H C3 3

2

i. O NHii. Zn,H O

R S

Chemistry84

14. The pair(s) of ions where BOTH the ions are precipitated

upon passing 2H S gas in presence of dilute HCl, is(are)

a. 2 2Ba , Zn b. 3 3Bi , Fe

c. 2 2Cu , Pb

d. 2 3Hg , Bi

15. Under hydrolytic conditions, the compounds used for preparation of linear polymer and for chain termination, respectively, are

a. 3 3CH SiCl and 3 4Si(CH )

b. 3 2 2(CH ) SiCl and 3 3(CH ) SiCl

c. 3 2 2(CH ) SiCl and 3 3 3(CH ) SiCl d. 4SiCl and 3 3(CH ) SiCl

16. When 2O is adsorbed on a metallic surface, electron

transfer occurs from the metal to 2O . The TRUE

statement(s) regarding this adsorption is(are)

a. 2O is physisorbed

b. heat is released

c. occupancy of *2p of 2O is increased

d. bond length of 2O is increased


This section contains TWO questions. Based on each paragraph, there will be TWO questions Each question has FOUR options (A), (B), (C) and (D). ONE OR

MORE THAN ONE of these four option(s) is(are) correct. For each question, darken the bubble(s) corresponding to all the

correct option(s) in the ORS. Marking scheme: +4 If only the bubble(s) corresponding to all the correct option(s)

is(are) darkened. 0 If none of the bubbles is darkened –2 In all other cases

Paragraph for Question Nos. 17 to 18

The hydrogen-like species 2+Li is in a spherically symmetric

state 1S with one radial node. Upon absorbing light the ion

undergoes transition to a state 2S . The state 2S has one radial

node and its energy is equal to the ground state energy of the hydrogen atom.

17. Energy of the state 1S in units of the hydrogen atom

ground state energy is : a. 0.75 b. 1.50 c. 2.25 d. 4.50 18. The orbital angular momentum quantum number of the

state 2S is :

a. 0 b. 1 c. 2 d. 3

Paragraph for Question No. 19 to 20 In the following reactions

19. Compound X is

a.

b.

c. d.

20. The major compound Y is

a.

b.

c. d.

CH3

CH3

CH3

CH2

CH3CH3

CHOOH

CH3

OH

CH3

O

2i. EtMgBr, H O8 8 ii. H , heat

C H O Y

2

4 2 4

H O

HgSO , H SO

4 2 2

2 2 2 2

Pd BaSO i. B H8 6 8 8H ii. H O . NaOH, H O

C H C H X


85Mock Test-1

ANSWER & SOLUTIONS

JEE-Main

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

a a c a d b b b d a

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

c d a b d d b c a a

21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

b d d c b a a a c b

1. (a)

2. (a) For BCC unit 6 cell, 3a 4r=

3 3

r a 4.294 4

= = × Å 1.85= Å

3. (c) Energy in 1st excited state 3.4 eV= −

4. (a) eff3 30

A

Z Mw

a 10 Nρ

−

×=

× × [For fcc, effZ 4= unit cell]

∴ 3 30

A

eff

a 10 NM

Z

ρ −× × ×

=w

–3 3 30 3 23(10.0 g cm (200pm) 10 cm 6 10 atoms)

4

−× × × ×

=

Thus, 112gmol− contains AN= atoms 236 10= × atoms

∴ 100 g contains 23

246 10100 5 10 atoms

12

×= × = ×

5. (d) n2NO NORx

G 2 G 2 G∆ = ∆ − ∆

2

1NO NO Rx2G G G∆ = ∆ + ∆

1

NO e p2G ( RT ln K )= ∆ + −

120.5[2 86600 R(298) n (1.6 10 )]= × − ×l

6. (b)

01

solute01 2

nP PX

P n n

−= =

+

185 183 2 1.2 / M

1.2 100185 185M 58

−= =

+

⇒ M 64=

7. (b) e cG RT ln K∆ = −

e c2494.2 8.314 300 ln K= × ⇒ 1cK e−=

1c

1K e 0.36

2.718−

= = =

12

2 2

2(B)(C)Q 4

[A] [1/ 2]

×= = =

cQ K ,> i.e. backward reaction.

8. (b) 2Cu 2e Cu (s)+ −+ →

2 mol 1 mol 63.5=

9. (d) min

10. (a) Amount of acetic acid adsorbed

33(0.06 0.042) 50 10 60

16 10 18 mg3

−−− × × ×

= = × =

11. (c) Ionic Radii order: 3 2N O F− − −> >

12. (d)

13. (a) It can act as an oxidising as well as reducing agent.

14. (b) 4BaSO is least soluble, 4BeSO is most soluble.

15. (d) The interhalogen compounds are generally more reactive

than halogens (except F2).

16. (d) Xe has the highest boiling point.

17. (b) No. of Geometrical isomers of

3 2[Pt(Cl)(py)(NH )(NH OH)] 3+ =

18. (c)

19. (a) % of Bromine 80 141 mg

100 24%188 250 mg

= × × =

20. (a) Only amines undergo acetylation and not acid amides.

21. (b)

t 40=

2.303 ak log

t a x=

−

2.303 ak log

40 a 0.875 a=

−

2.303 alog

40 0.125 a=

2.303log8 0.051 min

40= =

1/ 2

0.693 0.693t 13.58 min

k 0.051= = =

3CH

3CH

3CH

3CH

CHOO

3

2

O

Zn / H O→

5-keto-2-methyl hexanal

C

NH

2NH

C3CH

3CH COOH+

O

O

C

2NH

2NH

C3H C O

COH

OO

Chemistry86

22. (d) R I AgF R F AgI (Swarts Reaction)

23. (d)

24. (c)

25. (b) Glyptal is used in the manufacture of paints and lacquers.

26. (a) Vitamin ‘B’ and ‘C’ are water soluble.

27. (a) 2 6Zn [Fe(CN) ] is bluish white ppt.

28. (a) A1; B3, 4; C2; D3

(A) Inversion of cane sugar is pseudo first order reaction. (B) Hydrolysis of ester in the presence of acid is first order

while in the presence of base is second order reaction. (C) Photochemical reactions are of zero order. (D) SN2 reactions are of second order.

29. (c) Enthalpy is zero but entropy is not zero. Vibrational motion exists even at absolute zero.

30. (b) Molecularity of a reaction can be defined only for an elementary reaction because complex reaction does not take place in one single step and it is almost impossible for all the total molecules of the reactants to be in a state of encounter simultaneously.


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

1 2 9 8 4 4 3 4 a,b,d b,d

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

a d b,d c a,b,c a,b,c a,b b a a

1. (1) f fT iK m

0.0558 i 1.86 0.01 i 3

3 5 2Complex is [Co(NH ) Cl]Cl .

2. (2) Bridging does not allow the other 2 variants to exist.

Total no. of stereoisomers of M = 2

3. (9)

N is

4. (8)

5. (4) 3 36 6[Fe(SCN) ] and [Fe(CN) ]

In both the cases the electronic configuration of 3Fe will

be 2 2 6 2 6 51s , 2s , 2p ,3s ,3p ,3d .

Since SCN is a weak field ligand and CN is a strong field

ligand, the pairing will occur only in case of 36[Fe(CN) ] .

Case – 1 n(n 2) 5(5 2) 35 5.91 BM Case – 2 n(n 2) 1(1 2) 3 1.73 BM

Difference in spin only magnetic moment 5.91 1.73 4.18 4

6. (4) 2 3 2 2 3 2 2 3 2BeCl , N , N O, NO ,O ,SCl , ICl , I , XeF

5Case-1 3d (high spin) (no pairing)

Case-2(low spin)(pairing)

+:O N–– N

:O

: :O O N O :O N

Total no. of lone pairs = 8

–O O O

1 2 3

O O

4 5 6

–O

OO

7 8 9

–O

–O

HONaOH N

3CH

O

3H C

3H C

N Cl

CH3

NH3

CH3

CuCN / KCN2N 2NaNO / HCl

0 5 C

CN

CH3

3CH

Toluene

COOH

Benzoic acid

COCl

Benzoyl chloride

CHO

Benzaldehyde

2

4

H / PdBaSO

2SOCl2KMnO

87Mock Test-1

2BeC sp linear

3N sp linear

2N O sp linear

2NO sp linear

23O sp linear

3

2SCl sp linear

33I sp d linear

3

2ICl sp d linear

32XeF sp d linear

So, among the following only four (4) has linear shape and no

d-orbital is involved in hybridization.

7. (3) Single electron species don’t follow the (n ) rule but

multi electron species do.

Ground state of 2H 1s

First excited state of 1 1H 1s ,2s Second excited state of 1 0 1H 1s , 2s , 2p

8. (4) X Y 0rG 193KJ / mol

3M M 2e 0E 0.25V

0G for the this reaction is

0 0G nFE 2 ( 0.25) 96500 48250 J / mol

48.25 KJ/mol

So the number of moles of M oxidized using

X Y will be

193

4 moles48.25

9. (a, b, d) 3 2 46I ClO 6H SO

4 2 2Cl 6HSO 3I 3H O

10. (b, d)

(1)

(2)

(3)

(4)

11. (a)

12. (d)

13. (b, d) (a) 3 3H BO is a weak monobasic Lewis acid.

3 3 4H BO H OH B(OH) H . . .(i)

(b) Equilibrium (i) is shifted in forward direction by the

addition of syn-diols like ethylene glycol which forms a

stable complex with 4B(OH) .

3

3

H C

H C2C CH — CH — Br:Br

2C CH — HC

3

3

H C

H C

3

3 2

|— C — CH CH

CH

H C

2

3

H Br2

CH|

H C C — CH CH

O+

2H ,H O/Δ

3CH

O3CH

OH

O:OH

3CH2CH

O O

3CHCH

O

2H O

2H /Pt2 2 3H C CH —C —CH —CH

BrH3 2 2 3H C —CH —C—CH —CH

Optically inactiveBrH

2H /Pt2 3H C C —C — CH

BrH2H C


H

2H C

3 3H C — C — C — CH

2H /Pt3 2 3H C CH —C — CH — CH

BrH


3 2 3H C — CH — C — CH — CH

3H C — CH CH — C3CH

BrH3CH

BrHOptically active

2H /Pt3 2 2H C — CH — CH — C

(3 degenerate orbitals)

x y zP P P

Chemistry88

(c) It has a planar sheet like structure due to hydrogen

bonding. (d) H3BO3 is a weak electrolyte in water.

14. (c)

N2NH N

23 3NaNO / HCl

2 2

3 3

H HCH—CH —CH CH—CH —HC

H H

OH :OH| |C O C OC C

C C

2

3

2 H O

3

H CCH—CH —CH—COOH

H C 3

2

3

H CCH—CH —HC

H COH

OH

C O==

15. (a, b, c)

16. (a, b, c) Since container is thermally insulated so, q = 0, and it is a case of free expansion therefore W = 0 and

E 0. So, 1 2T T . Also, 1 1 2 2P V P V .

17. (a, b) 32 22Fe H O 2OH 2

2 22Fe 2H O O

2 2 2Na O H O 2 2H O NaOH

18. (b) 2 2N (g) 2H (g)

2 2 3N (g) 2H (g) 2NH (g); H 0

Increasing the temperature lowers equilibrium yield of ammonia. However, at higher temperature the initial rate of

forward reaction would be greater than at lower temperature that is why the percentage yield of NH3 too would be more initially.

19. (a) →1; B→3; C→4; D→2

(A)

(B)

(C)

3 3Ph CH — CO — Ph CH

(D)

26 5 3 6 5 2C H CO CH O C H CO

20. (a)→2; B→3; C→1; D→4

(A)

(B)

(C)

(D)

d-d antibonding

p-d antibonding

p-d bonding

d-d bonding

O

6 5C H

OO 3CH

26 5 3 3

3

O|

C H — CO CH C — CH|

CH

3CHO

6 5C H

OO

3CH

6 5C H

2 3 3Ph — C H CH — CO — CH

6 5 2 2 2 3

3

O|

C H — C H CO Ph — CH — C — CH|CH

3CHO

5 6 2H C CH

OO

3CH

2 6 5CH C H

6 5 2 2 2C H CH CO CH O

3CH

O

5 6 2H C CH

OO

OH

OH

OHBr

Br

ClCl

3CMe

I

3CMe

3CMe

Cl

3CMe

OH

2I

Rxn(i)

2Br

Rxn(ii)

2Cl

Rxn(iii)

O O

OO

24H O

(stable complex)

B

–

OO O

O

OO

OO H

H

HH

HH

HH

B

89Mock Test-1


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

8 4 4 75 5 6 3 9 a a

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

b a b,c c,d b b,c,d a b c d

1. (8) 2 22 4 2 4[Fe(C O )(H O)] MnO 8H

2 32 2Mn Fe 4CO 6H O

So, the ratio of rate of change of

[H ] to that of rate of change of 4[MnO ] is 8.

2. (4)

3. (4)

I

II

III

IV

4. (75)

Z 1 [1

th8

is shared in first layer and

1

2is shared in second layer

1

Z 4 0.5 18

2 2 2(2 ) r h x 2h rfi =

2 2 2Area 4 [ (2 ) 2 ]r a r a r= = fi =

Vol 24 2 r r 34 2 r

3

3

41

3D 0.75 75%4 2 3 2

r

r

5. (5) 2 2[Co(en) Cl ] will show cis-trans isomerism

32 2 4 2[CrCl (C O ) ] will show cis-trans isomerism

2 4 2[Fe(H O) (OH) ] will show cis-trans isomerism

4 3 2[Fe(CN) (NH ) ] will show cis-trans isomerism

22 3[Co(en) (NH )Cl] will show cis-trans isomerism

23 4 2[Co(NH ) (H O)Cl] will not show cis-trans

isomerism (although it will show geometrical isomerism)

6. (6) 2 6 3 2B H 6MeOH 2B(OMe) 6H

1 mole of 2 6B H reacts with 6 moles of MeOH to give 2

moles of 3B(OMe)

3 moles of 2 6B H will react with 18 moles of MeOH to

give 6 moles of 3B(OMe) .

7. (3) HX H X

[H ][X ]Ka

[HX]

HY H Y

[H ][Y ]Ka

[HY]

m for 1mHX m for

2mHY

1 2m m

1

10 2Ka C

1

2

m1 1 0

m1

Ka C

2

2

m2 2 0

m2

Ka C

2

1 1 1

2 2 2

Ka C m

Ka C m

2

0.01 10.001

0.1 10

1 2pKa pKa 3

8. (9) In conversion of 23892 U to 206

82 Pb, 8 - particles and 6β particles are ejected.

2r

h

x

2CO Me CHO

2

BIBAL HToluene, 78 C H O

COCl CHO2

4

HPd BaSO

2CHCl CHO2H O

100 C

CHO

3

CO, HClAnhydrous AlCl / CuCl

H

+

H

HO

HOHO

OH

OH

(Q)

(P)4aqueous dilute KMnO

(excess) 0 C

Chemistry90

The number of gaseous moles initially 1mol

The number of gaseous moles finally 1 8 mol; (1 mol

from air and 8 mol of 42 He ) So the ratio 9 /1 9

9. (a)

10. (a)

11. (b)

12. (a)

13. (b, c) H — O — Cl

(I) . .

. .H — O — Cl O

(II) . .

H — O — Cl O||O

(III)

O||

H — O — Cl O||O

(IV)

14. (c, d) 2 2 2 3Cu , Pb , Hg , Bi give ppt. with 2H S in presence

of dilute HCl.

15. (b)

16. (b, c, d) * Adsorption of 2O on metal surface is

exothermic.

* During electron transfer from metal to 2O electron

occupies *2p orbital of 2O .

* Due to electron transfer to 2O the bond order of 2O

decreases hence bond length increases.

For Question Nos. 17 to 18

17. (c) 1

21H

S 1H2

E 3 9E 2.25 E .

42

3 2Me SiCl, H O

3 3

n3 3

CH CH| |

Si — O — Si — O — Si ——O — Si| |

CH CH

MeMeMe

MeMe

Me

3 3

n3 3

CH CH| |

H — O — Si — O — Si ——O — H| |

CH CH

2

3

H O

3

CH|

Cl — Si — Cl|

CH

3

3

CH|

HO — Si — OH|

CH

N N Ph OH

(W)

3NaNO ,HCl β Napthol/NaOH0 C

(V)

2NH–

2N Cl

T

CH 3 3H C CH

3

3

CH|

C — CH|

OO

H U

2O

3 3H C — CH — CH

O||C — H3H C

2 2CH — CH — NH||OH

. .

3H C

(S)

N

OH

3H C NH

OH

22H O

O

3H C 2NH

2CHCH

OH

O||C — H3H C

2 3CH — CH — NH||O

3NH

3H C 3

2

(i) O(ii) Zn.H O

3H C

R

O||C — H

2CH — C — H||O

3

H

CHC C2Ph CH

H

trans

3CH

HC C2Ph CH

H

cis

91Mock Test-1

18. (b)

2S is 3p l = 1


8 6C H double bond equivalent

6

8 1 62

3

H / heat3 3

OH CH| |

Ph — C — CH Ph — C CH — CH| (Y)Et

19. (c) 20. (d)

2 2CH CH OH

2

(i) EtMgBr(ii) H O

(X)

3

O||C — CH

C CH

2CH CH

4 2 4 2HgSO , H SO , H O

2 8

2 2 2

(1) B H(2) H O , NaOH, H O

Chemistry92


1. 2 g of oxygen contain same number of atoms as contained by

a. 0.5 g hydrogen b. 4.0 g sulphur c. 7.0 g nitrogen d. 2.3 g sodium

2. The axial angles in triclinic crystal system are

a. 90 b. 90 , 90

c. 90 d. 90

3. A sample of drinking water was found to be severely

contaminated with chloroform supposed to be a

carcinogen. The level of contamination was 15 ppm (by mass). Determine the molality of chloroform in the water sample.

a. b.

c. d.

4. The temperature at which hydrogen molecules will have the same root mean square velocity as oxygen molecules

have at 27 C is

a. 25 C b. 7.93 C

c. 248 C d. 127 C

5. An electron makes a transition from orbit n = 4 to the orbit n = 2 of a hydrogen atom. The wave number of the emitted radiations (R = Rydberg's constant) will be

a. 16

3 R b.

2 R

16 c.

3 R

16 d.

4 R

16

6. Select the compound in which chlorine is assigned the oxidation number +5

a. 4HClO b. 2HClO c. 3HClO d. HCl

7. The molar conductivity is maximum for the solution of concentration

a. 0.001 M b. 0.005 M c. 0.002 M d. 0.004 M

8. The number and type of bonds between two carbon atoms in calcium carbide are

a. One sigma, one pi b. One sigma, two pi c.Two sigma, one pi d. Two singma, two pi

9. The standard state gibbs free energy change for the given

isomerization reaction cis-2-pentene trans-2-pentene

is at If more trans-2-pentene is added

to the reaction vessel, then

a. More cis-2-pentene is formed b. Equilibrium is shifted in the forward direction

c. Equilibrium remains unaffected d. Additional trans-2-pentene is formed

10. The dissociation constant of is

The hydrolysis constant for 0.1 M sodium acetate is

a. b.

c. d.

11. The half-life of a first order reaction having rate constant

is

a. 12.1 h b. 9.7 h c. 11.3 h d. 1.8 h

12. Which characteristic is true in respect of colloidal particle a. They always have two phases

b. They are only in liquid state c. They can't be electrolysed

d. They are only hydrophilic

13. The heat of transition of graphite into diamond

would be, where

a. b.

c. d. None of these

14. All the nuclei from the initial element to the final element

constitute a series which is called a. g-series b. b-series

c. b-g series d. Disintegration series

15. The following compound will undergo electrophilic substitution more readily than benzene

a. Nitrobenzene b. Benzoic acid

c. Benzaldehyde d. Phenol

16. Cis and trans 2-butene are a. Conformational isomers

b. Optical isomers

c. Position isomers d. Geometrical isomers

17. Electrolysis of a concentrated solution of sodium fumarate

gives

a. Ethylene b. Ethane c. Acetylene d. Vinyl alcohol

3CHCl ,

4 12.12 10 mol kg 4 11.26 10 mol kg 4 10.12 10 mol kg 4 15.36 10 mol kg

–3.67kJ / mol 400K.

3CH COOH 51.8 10 .

45.56 10 105.56 1051.8 10 91.8 10

5 1K 1.7 10 s

t( H )

2 2C(graphite) O (g) CO (g); H x kJ

2 2C(diamond) C (g) CO (g) ; H y kJ 1(x y) kJ mol 1(x y) kJ mol1(y x) kJ mol

93Mock Test-2

18. Identify and in the following sequence

a.

b.

c.

d.

19. Consider the following alcohols

(i) 1-Phenyl-1-propanol

(ii) 3-Phenyl-1-propanol

(iii) 1-Phenyl-2-propanol

The correct sequence of increasing order of reactivity of

these alcohol in their reaction with HBr is

a. (i), (ii), (iii) b. (ii), (i), (iii)

c. (i), (iii), (ii) d. (ii), (iii), (i)

20. Reaction between diethyl cadmium and acetyl chloride

leads to the formation of

a. dimethyl ketone b. ethylmethyl ketone

c. diethyl ketone d. acetaldehyde

21. Which of the following is basic?

a. b.

c. d.

22. The compound X is:

a. b.

c. d.

23. Which of the following is a synthetic polymer

a. Rubber b. Perspex

c. Protein d. Cellulose

24. The correct statement in respect of protein haemoglobin is that it

a. Acts as an oxygen carrier in the blood b. Forms antibodies and offers resistance to diseases c. Functions as a catalyst for biological reactions d. Maintains blood sugar level

25. Which one of the following is known as broad spectrum antibiotics

a. Streptomycine b. Ampicillin c. Chloramphenicol d. Penicillin G

26. The element or elements whose position is anomalous in the periodic table is

a. Halogens b. and Ni

c. Inert gases d. Hydrogen

27. Flux added in the extraction of iron is a. Silica b. Felspar c. Limestone d. Flint

28. The correct order of the increasing ionic character is

a.

b.

c.

d.

29. Nesseler's reagent is

a. 2 4K HgI b. 2 4K HgI KOH

c. 2 2K HgI KOH d. 2 4K HgI Hg

30. The primary valence of the metal ion in the co-ordination

compound 2 4K Ni CN is

a. Four b. Zero c. Two d. Six

X YX

2 5C H Br Yproduct 3 7C H 2NH

4X KCN,Y LiAlH

3X KCN,Y H O

3 3X CH Cl,Y AlCl / HCl

3 2 2X CH NH ,Y HNO

3 2CH CH OH 2 2H O

2 2HOCH CH OH 3CH COOH

2 5Na, C H OH3CH CN X.

3 2 2CH CH NO 3 2CH CH COOH

6 5 3 2C H N(CH ) 6 5 2C H CONH

Fe, Co

2 2 2 2BeCl MgCl CaCl BaCl

2 2 2 2BeCl MgCl BaCl CaCl

2 2 2 2BeCl BaCl MgCl CaCl

2 2 2 2BaCl CaCl MgCl BeCl


Chemistry94

JEE ADVANCE PAPER-I

SECTION 1 Contains 8 Questions

The answer to each question is a single digit integer ranging from 0 to

9 (both inclusive).

1. At 400 K, the root mean square (rms) speed of a gas X

(molecular weight = 40) is equal to the most probable speed

of gas Y at 60 K. The molecular weight of the gas Y is

2. Based on VSEPR theory, the number of 90 degree

F−Br−F angles in is

3. 20% surface sites have adsorbed On heating gas

evolved from sites and were collected at 0.001 atm and

298 K in a container of volume is 2.46 cm3. Density of

surface sites is and surface area is 1000

cm2, find out the no. of surface sites occupied per

molecule of

4. The total number of contributing structure showing

hyperconjugation (involving C–H bonds) for the

following carbocation is

5. The volume (in mL) of 0.1 M required for

complete precipitation of chloride ions present in 30 mL

of 0.01 M solution of as silver chloride

is close to

6. Among and

the total number of black coloured sulphides is

7. 29.2% (w/w) HCl stock solution has a density of 1.25 g

The molecular weight of HCl is 36.5 g The

volume (mL) of stock solution required to prepare a 200

mL solution of 0.4 M HCl is

8. The maximum number of isomers (including

stereoisomers) that are possible on monochlorination of

the following compound is

SECTION 2 Contains 10 Multiple Choice Questions

With one or more than one correct option

9. Sodium metal crystallises in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately

a. 1.86 Å b. 3.22 Å c. 5.72 Å d. 0.93 Å

10. 18 g glucose is added to 178.2 g water. The vapor

pressure of water (in torr) for this aqueous solution is : a. 7.6 b. 76.0 c. 752.4 d. 759.0

11. For one mole of a van der Waals gas when b = 0 and T = 300 K, the PV vs. 1/V plot is shown below. The value of the van der Waals constant a (atm. liter2 mol2) is.

a. 1.0 b. 4.5 c. 1.5 d. 3.0

12. A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant

temperature of As it does so, it absorbs 208 J of

heat. The values of q and w for the process will be

a. b.

c. d.

13. The first ionisation potential of Na is 5.1 eV. The value of

electron gain enthalpy of will be

a. b.

c. d.

14. In a galvanic cell, the salt bridge a. does not participate chemically in the cell reaction. b. stops the diffusion of ions from one electrode to another. c. is necessary for the occurrence of the cell reaction d. ensures mixing of the two electrolytic solutions

15. Stability of the species of and increase in

the order of

a. b.

c. d.

5BrF

2N . 2N

14 26.023 10 / cm

2N .

3AgNO

2 5 2[Cr(H O) Cl]Cl ,

2 2 3PbS, CuS, HgS, MnS, Ag S, NiS,CoS, Bi S

2SnS ,

1mL . 1mol .

6 I2 6(C H P )

37.0 C.

(R 8.314 J / mol K) (ln 7.5 2.01)

q 208 J, w 208 J q 208 J, w 208 J

q 208 J, w 208 J q 208 J, w 208 J

Na

–2.55 eV 5.1eV

10.2 eV 2.55 eV

2 2Li , Li 2Li

2 2 2Li Li Li 2 2 2Li Li Li

2 2 2Li Li Li 2 2 2Li Li Li

(Graph not to scale)

0 2.0 3.0

20.121.623.124.6

PV

(li

ter-

atm

mol

–1)

11(mol liter )

V

3 2CH CH 2 3CH CH

C

H

3CH

3 2 3H C CH CH

95Mock Test-2

16. How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?

a. 0.1 L b. 0.9 L c. 2.0 L d. 9.0 L

17. NiCl2 exhibits temperature dependent

magnetic behavior (paramagnetic / diamagnetic). The coordination geometries of Ni2+ in the paramagnetic and diamagnetic states are respectively.

a. tetrahedral and tetrahedral b. square planar and square planar c. tetrahedral and square planar d. square planar and tetrahedral

18. Four successive members of the first row transition elements are listed below with atomic numbers. Which one of them is

expected to have the highest value?

a. b.

c. d.

SECTION 3 Contains 2 Match The Following Type Questions

You will have to match entries in Column I with the entries in Column II.

19. Match gases under specified conditions listed in Column-I with their properties/laws in Column-II.

Column I Column II

(A) Hydrogen gas

273K)

1. Compressibility

factor

(B) Hydrogen gas (P 0,

2. Attractive forces are dominant

(C) 3.

(D) Real gas with very large

molar volume

4.

a. A 2; B 3; C 2; D 1, 4

b. A 1, 3; B 1; C 2; D 4

c. A 3; B 4; C 3, 4; D 1, 2

d. A 1, 2; B 3; C 1, 2; D 1, 4

20. The standard reduction potential data at 25ºC is given

below:

Match of the redox pair in Column I with the values

given in Column II.

Column I Column II

(A) 1. – 0.18 V

(B) 2. – 0.4 V

(C) 3. – 0.04 V

(D) 4. – 0.83 V

a. A 2; B 3; C 1; D 4

b. A 1; B 2; C 3; D 4

c. A 3; B 4 ; C 1; D 2

d. A 2; B 3; C 1; D 4

2 5 2 6 5 2P(C H ) (C H )

3 2

0

M / ME

Cr(Z 24) Mn(Z 25)

Fe(Z 26) Co(Z 27)

(P 200 atm,

T 1

T 273K)

2CO (P 1 atm, T 273K) P nRT

P(V nb) nRT

3 2E (Fe , Fe ) 0.77 V; 2E (Fe , Fe) 0.44V 2E (Cu , Cu) 0.34V; E (Cu , Cu) 0.52V

2 2E [O (g) 4H 4e 2H O] 1.23V;

2 2E [O (g) 2H O 4e 4OH ] 0.40V

3E (Cr , Cr) 0.74V; 2E (Cr , Cr) 0.91 V

E

3E (Fe , Fe)

2E (4H O 4 4OH ) 2E (Cu Cu 2Cu )

3 2E (Cr , Cr )


Chemistry96


SECTION 1 Contains 8 Questions.


9 (both inclusive).

1. The maximum number of electrons that can have principal

quantum number, n = 3, and spin quantum number,

is

2. The dissociation constant of a substituted benzoic acid at 25ºC is 1.0 × 10–4. The pH of a 0.01M solution of its

sodium salt is

3. The total number of cyclic structural as well as stereo isomers possible for a compound with the molecular

formula is

4. To an evacuated vessel with movable piston under external pressure of 1 atm, 0.1 mol of He and 1.0 mol. of

an unknown compound (vapour pressure 0.68 atm. At 0ºC) are introduced. Considering the ideal gas behaviour,

the total volume (in litre) of the gases at 0ºC is close to

5. The coordination number of Al in the crystalline state of

is

6. Among the following, the number of compounds than can

react with to give is

7. If the freezing point of a 0.01 molal aqueous solution of a

cobalt (III) chloride-ammonia complex (which behaves as a strong electrolyte) is –0.0558ºC, the number of

chloride(s) in the coordination sphere of the complex is [Kf of water = 1.86 K kg mol–1]

8. A student performs a titration with different burettes and

finds titre values of 25.2 mL, 25.25 mL and 25.0 mL. The number of significant figures in the average titer value is



9. The correct statement(s) regarding defects in solids is(are)

a. Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion

b. Frenkel defect is a dislocation defect c. Trapping of an electron in the lattice leads to the

formation of F-centre

d. Schottky defects have no effect on the physical properties of solids

10. Mixture(s) showing positive deviation from Raoult’s law at 35ºC is(are)

a. carbon, tetrahedral + methanol b. carbon disulphide + acetone c. benzene + toluene d. phenol + aniline

11. For a gaseous state, if most probable speed is denoted by

C*, average speed by and mean square speed by C,

then for a large number of molecules the ratio of these speeds are

a.

b.

c.

d.

12. The species having bond order different from that in CO is a. NO– b. NO+ c. CN– d. N2

13. Which of the following is the wrong statement?

a. ONCF and are not isoelectronic

b. molecules is bent

c. Ozone is violet–black in solid state d. Ozone is diamagnetic gas

14. The qualitative sketches I, II and III given below show the variation of surface tension with molar concentration of

three different aqueous solutions of KCl, 3CH OH and

3 2 11 3CH (CH ) OSO Na at room temperature. The correct

assignment of the sketches is

a. I : KCl II : 3CH OH III : 3 2 11 3CH (CH ) OSO Na

b. I : 3 2 11 3CH (CH ) OSO Na II : 3CH OH III : KCl

c. I : KCl II : 3 2 11 3CH (CH ) OSO Na III : 3CH OH

d. I : 3CH OH II : KCl III: 3 2 11 3CH (CH ) OSO Na

15. Using the data provided, calculate the multiple bond

energy (kJ mol–1) of CC bond in C2H2. That energy is

(take the bond energy of a C—H bond as 350 kJ mol–1)

s

1m ,

2

5 10C H

3AlCl

5PCl 3POCl

2 2 2 2 2 4 4 10O , CO , SO , H O, H SO , P O

C

C* : C : C 1.225 :1.128 :1

C*: C : C 1.128 :1.1225 :1

C* : C : C 1:1.228 :1.225

C* : C : C 1:1.225 :1.128

ONO

3O

Concentration

Sur

face

tens

ion II

Concentration

Sur

face

tens

ion

I

Concentration

Sur

face

tens

ion III

97Mock Test-2

22C(s) + H (g) ;

a. 1165 b. 837 c. 865 d. 815

16. A solution of (–)-1-chloro-1-phenylethane in toluene racemises slowly in the presence of a small amount of

due to the formation of

a. carbanion b. carbene c. carbocation d. free radical

SECTION 3 Contains 2 Paragraph Type Questions

Each paragraph describes an experiment, a situation or a problem.

Two multiple choice questions will be asked based on this paragraph.

One or more than one option can be correct.


The electrochemical cell shown below is a concentration cell.

(saturated solution of sparingly soluble salt,

M. The emf of the cell depends

on the difference in concentration of ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.

17. The value of for the given cell is

(take 1 F = 96500 C

a. –5.7 b. 5.7 c. 11.4 d. –11.4

18. The solubility product of MX2 at 298 K

based on the information available for the given concentration cell is (take 2.303 × R × 298 / F = 0.059 V)

a. b.

c. d.


Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately

are present in a few grams of any chemical

compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution of NaCl is prepared and 500mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass:

1 Faraday coulombs)

19. The total number of moles of chlorine gas evolved is a. 0.5 b. 1.0 c. 2.0 d. 3.0

20. If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is

a. 200 b. 225 c. 400 d. 446

2 2C H (g) 1H 225 kJ mol

2C(s) 2C(g)1H 1410 kJ mol

2H (g) 2H(g)1H 330 kJ mol

5SbCl ,

2M|M 2 3

2MX ) || M (0.001 mol dm ) |

2M

1G (kJ mol )1mol )

3 9sp(K ;mol dm )

151 10 154 10121 10 124 10

236.023 10 )

Na 23, Hg 200, 96500


Chemistry98

ANSWER & SOLUTIONS

JEE-Main

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

b c b c c c a b a b

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

c a b d d d c a d b

21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

a a b a c d c a b c

1. (b) Number of atoms in 2 g of oxygen

232 6 10

16

× ×=

(1) 23 236 10 0.5 3 10× × = ×

(2)

23226 10 4

7.5 1032

× ×= ×

(3)

23236 10 7

1.5 1028

× ×= ×

(4)

23226 10 2.3

6 1023

× ×= ×

∴

The correct answer is (b).

2. (c) The axial angles in triclinic crystal system are different

and none is perpendicular to any of the others i.e.,

90 .α β γ≠ ≠ ≠ °

3. (b) Level of contamination = 15 ppm

= 15 parts in parts Mass % of

Thus, the water sample contains

(by mass) of

Amount of

Mass of water (solvent) =

So, molality of in solution

4. (c) r.m.s. 2

3RT(H )

2=

r.m.s. 2

3R 400(O )

32

×=

( T 273 127 400 K= + = )

3 RT 3R 300

2 32

×=

⇒ T 400

2 32=

∴ 400 2

T 25 K32

×= = or T 25 273 248 C= − = − °

5. (c) Wave number2 2

1 2

1 1 1 1 1 3RR R

4 16 16n nλ

= − = − =

6. (c)

⇒ ;

7. (a) Since molar conductance

8. (b) and

9. (a) Equilibrium shifts backward by Le-chatelier’s

principle.

10. (b)

11. (c)

12. (a) Dispersion medium and dispersed phase are phases of

colloid.

13. (b) Graphite diamond

14. (d) Definition of disintegration series.

15. (d) Phenol will undergo electrophilic substitution more

readily than benzene.

16. (d) Cis and trans 2-butene are geometrical isomers.

17. (c)

18. (a)

19. (d) HBr reacts with alcohols through the formation of

carbocation. The stable is the carbocation formed, more is

the reactivity of alcohol with HBr. The carbocation

formed are

(1) (2) (3)

610

3CHCl

3

6

15100 1.5 10

10

−= × = ×

31.5 10 %

−× 3CHCl

3 1

15gCHCl 0.126mol

119.5g mol−= =

6 310 g 10 kg=

3CHCl

4 1

3

0.126mol1.26 10 molkg

10 kg

− −= = ×

3HClO

∗

1 2 3 0+ − × =x 6 1 5= − = +x

1

Molarity∝

σ1 π2

1410w

b 5

a

K 1 10K 5.56 10

K 1.8 10

−−

−

×= = = ×

×

5 1K 1.7 10 s

− −= ×

5

1/ 2

0.693 0.693t 10 11.32h

K 1.7= = × =

→ 1

tH ( ) kJ mol .−∆ = −x y

CHCOONa||CHCOONa

Electrolysis

22H O+ →

Acetylene

CH|||CH

2 22CO 2KOH H+ + +

4LiAlH (Y)KCN(X)

2 5 2 5C H Br C H CN→ →

2 5 2 2 3 7 2C H CH NH (C H NH )

4X KCN,Y LiAlH= =

2 3CH CH CH+− −2 2 2CH CH CH+

2 3CH CH CH+

99Mock Test-2

20. (b)

21. (a) Ethanol is the weakest acid among these, hence it is

most basic.

22. (a)

23. (b) Perspex is a synthesised polymer.

24. (a) Four ions of each haemoglobin can bind with 4

molecules of and it is carried as oxyhaemoglobin.

25. (c) Chloramphenicol is broad spectrum antibiotic used in the treatment of typhoid, dysentry, acute fever.

26. (d) It shows similarities with both alkali metals as well as halogens.

27. (c) Impurities of 2SiO is present in iron ore so basic flux

3CaCO is added.

2 3Flux Impurity Slag

CaO SiO CaSiO

28. (a)

As we go down the group I.E. decreases. Hence ionic character increases.

29. (b) 2 2 4

Nessler's reagent

2KI HgI K Hgl KOH

30. (c) Primary valencies are also known as oxidation state.

2 4K [ Ni(CN) ], 2 x 4 0 x 2


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

4 0 2 6 6 6 8 8 a c

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

c a b a b d c d d c

1. (4)

,

2. (0)

All four planar bonds (F−Br−F) will reduce from 90° to

84.8° after bp repulsion.

So among the following only four (4) has linear shape and no d-orbital is involved in hybridization.

3. (2)

By ideal gas,

Now molecules of

Now total surface sites available

Surface site used to adsorb

Sites occupied per molecule of

4. (6)

5. (6) Number of ionisable in is 2

Millimoles of

Millimoles of required

6. (6) Black coloured sulphides

* Bi2S3 in its crystalline form is dark brown but

precipitate obtained is black in colour.

7. (8) Molarity

Cd2 3

2 3

CH CH

CH CH32CH COCl

3 2 3 22CH COCH CH CaCl

2 5Na, C H OH3 3 2 2CH CN 4H CH CH NH (X)

2Fe

2O

4 2 4 8Hb 4O Hb O

2 2 2 2BeCl MgCl CaCl BaCl

(Xgas )(400K) (Ygas )(60K )rms mpV V

M.W.(X gas) 40;M.W.(Ygas) x

1 1

1 2

3RT 2RT

M M

400 3 2 60

40 x

12030

x x 4

p

2

2NP 0.001atm, T 298 K, V 2.46 cm

PV nRT

2

37

N

PV 0.001 2.46 10n 1.0 10

RT 0.0821 298

23 7 162N 6.023 10 1 10 6.023 10

14 176.023 10 1000 6.023 10

17 162

20N 6.023 10 12.04 10

100

16

2 16

12.04 10N 2

6.02 10

6 H atoms are there

Cl 2 5 2[Cr(H O) Cl]Cl

Cl 30 0.01 2 0.6

Ag 0.6

0.6 0.1 V

V 6 ml

2PbS,CuS, HgS, Ag S, NiS,CoS

2 3Bi S

solnsol

w w sol sol w

W dn W W 100 d1000

V M v M W W M 100

w

W% d 10

WM

BrF F

F FF

Chemistry100

⇒

8. (8)

Total = 2 + 4 + 1 + 1 = 8

9. (a) For BCC 6 unit cell,

Å Å

10. (c) Moles of glucose =

Moles of water =

⇒

torr

= 752.4 torr

11. (c)

fi Slope = – a

Slope

12. (a) (as it absorb heat)

13. (b) E.A = Ionisation potential

EA of

14. (a)

15. (b) B.O. of , B.O. of

Hence stability order

16. (d) ⇒

17. (c)

18. (d) Factual

19. (d) A → 1, 2 ; B→ 3 ; C→ 1, 2 ; D → 1, 4

(A) at high pressure and low temperature.

Equation reduces to

29.2 1.25 1010M

36.5

× ×= = 1 1 2 2M V M V=

⇒ 110 V 0.4 200× = ×

0.4 200V 8ml

10

×= =

3a 4r=

3 3r a 4.29

4 4= = × 1.85=

180.1

180=

178.29.9

18=

⇒ Totaln 10=

⇒P 0.1

P 10

∆=

°

P 0.01P∆ = °

0.01 760 7.6= × =

SP 760 7.6= −

b 0,T 300K, n 1= = =

2

2

anP (V nb) RT

V

+ − =

2

aP (V) RT

V

+ =

⇒a

PV RTV

+ =

1PV a RT

V

= − × +

y mx C= +

2 1

2 1

y y 20.1 21.61.5

x x 3 2

− −= =

− −

q 208 J,= +

2Re w 10

1

vw 2.303 nRT log

v

= −

10

3752.303 (0.04 8.314 (310) log 208 J

50

= − × × × = −

∴ Na 5.1eV+ =−

2Li 0.5+ = 2Li 0.5− =

2 2 2Li Li Li− += < <

0.1 1 (1 v) 0.01× = + ×0.1

1 v0.01

+ =

⇒ 1 v 10+ = ⇒ v 10 1 9L= − =

22 2 5 2 6 5 2[NiCl P(C H ) (C H ) ]+

2 8Ni (28) : [Ar]4s 3d+ °

mPVZ

RT=

2

2

anP (V nb) nRT

V

+ − =

P(V nb) nRT.− =

Hybridisation is dsp2 (diamagnetic)

Square planar

Hybridisation is sp3 (paramagnetic)

tetrahedral

3d8 In strong ligand

In weak ligand 3d8

3 2 2 3CH CH C CH CH 1−

2CH Cl

H

*

3 2 2 3CH CH C CH CH 1−

3CH

H

3 2 3CH CH C CHCH Two Enantiomeric pairs 4− =

3CH

H

*

*

3CH

Cl

Br

Br

Cl

3CH

101Mock Test-2

(B) For hydrogen gas value of Z = 1 at P = 0 and it increases continuously on increasing pressure.

(C) CO2 molecules have larger attractive forces, under normal conditions.

(D) at very large molar volume Z 1.

20. (c) A 3; B 4 ; C 1 ; D 2

(A)

(B) . . .(i)

. . .(ii)

So . . .(iii)

For IIIrd reduction

(C)

(D)


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

9 8 7 7 6 5 1 3 b, c a, b

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

c a a d d c d b b d

1. (9) Number of orbital for is

Number of electrons for and

2. (8) ) = 1 × 10–4

(1–h 1)

3. (7) Cyclic

For 3rd structure 2 cis-trans and 1 optical isomer are

possible. Total 7 isomers.

4. (7) Let unknown is X.

Now

=7

5. (6) Coordination number of Al is 6. It exists in ccp lattice with 6 coordinate layer structure.

6. (5)

7. (1) f fT = iK m

8. (3)

9. (b, c)

10. (a, b)

(a) 4 3CCl CH OH Positive deviation from Raoult’s

law

(b) 2CS

Positive deviation from Raoult’s law

(c) 6 6 7 8C H C H Ideal solution

(d)

Negative deviation from Raoult’s law.

11. (c)

1 : 1.128 : 1.225

12. (a) (16 electron system)

Bond order =2.

mPVZ ,

RT

3 3 2 2

o o o

Fe / Fe Fe / Fe Fe / FeG G G

3 3 2 2

o o o

(Fe / Fe) (Fe / Fe ) Fe / Fe3 FE 1 FE ( 2 FE )

3

o

Fe / FeE 0.04 V

2 2O (g) 2H O 4e 4OH E 0.40 V

2 22H O O (g) 4H 4e E 1.23 V

24H O 4H 4OH E 0.40 1.23 0.83 V.

2 2

o o o

(Cu / Cu) Cu / Cu Cu / CuG 1 FE ( 1 F E )

2 2

o o o

Cu / Cu Cu / Cu Cu / Cu2 FE 1 FE ( 1 F E )

2

o

Cu / CuE 0.18 V.

3 2 3 2

o o o

Cr / Cr Cr / Cr Cr / CrG G G

3 2 3 2

o o o

Cr / Cr Cr / Cr Cr / Cr1 F E 3 F E ( 2 F E )

3 2

o

Cr / CrE 0.4 V.

n 3 2n 9

n 3 s

1m 9

2

a 6 5K (C H COOH

6 5pH of 0.01MC H COONa

16 5 2 6 5

0.01h0.01(1 h) 0.01h

C H COO H C H COOH OH

2w

ha

K 0.01hK

K 1 h

14 2 2

4

10 10 h

10 1 h

4 6[OH ] 0.01h 0.01 10 10 8[H ] 10 pH 8

5 10C H

He total xp p P (1 0.68) atm 0.32 atm

He He

RTp n

V

He

RT 0.10 0.082 273v

p 0.32

0.0558 i 1.86 0.01 i 3

3 5 2Complex is [Co(NH ) Cl]Cl

2RT 8RT 3RTC*: C : C

M M M

82 : 3

3.14

NO

+

OH 2NH

3CH 3CH

O

C

Chemistry102

and N2 are isoelectronic with CO therefore all

have same bond order (=3).

13. (a) ONCF and ONO– are isoelectronic in nature.

14. (d) Impurities affect surface tension appreciably. It is observed that impurities which tend to concentrate on surface of liquids, compared to its bulk lower the surface tension.

Substances like detergents, soaps 3 2 11 3[CH (CH ) SO Na ]

decreases the surface tension sharply.

Those like alcohol (e.g., 3 2 5–CH OH, C H OH) lower the

surface tension slightly. This can also be related to the fact

that 3CH OH has smaller dielectric constant. Dielectric

constant is directly proportional to surface tension. So, on

adding 3CH OH in water, overall dielectric constant

decreases and surface tension decreases. Inorganic impurities present in bulk of a liquid such as

KCl tend to increase the surface tension of water.

15. (d)

BE(H2) + BE(H2) + Hsub (C) – BE (C – H) × 2 +

BE(C C ) Hrxn

x= 815

16. (c) Racemises slowly due to formation of intermediate carbocation.


17. (d)

18. (b)

5 3 154(10 ) 4 10- -= = ¥


19. (b)

At anode:

Moles of in 500 ml.

Therefore 1 mole of Cl2 evolves.

20. (d) Na—Hg (amalgam) formed = 2 moles at cathode.

NO , CN

2 2 22C(s) H (g) C H (g)(H C C H)

330 1410 [350 2 x] 225

cellG nFE

2 96500 0.059 111.387 kJ mol

11.4 kJ

10 3

0.0591 CE log

2 10

10 3

0.0591 C0.059 log

2 10

23

C10

10

2 5C [M ] 10 M.

2 2 3spK [M ][X ] 4s

NaCl Na Cl

22Cl Cl

Cl 2

103Mock Test-3


1. The volume of 1.0 g of hydrogen in litres at N.T.P. is a. 2.24 b. 22.4 c. 1.12 d. 11.2

2. Which set of characteristics of ZnS crystal is correct? a. Coordination number (4: 4) : ccp; Zn2+ ion in the alternate

tetrahedral voids b. Coordination number (6: 6); hcp; Zn2+ ion in all tetrahedral

voids c. Coordination number (6: 4) : hcp; Zn2+ ion in all octahedral

voids d. Coordination number (4: 4); ccp; Zn2+

ion in all tetrahedral voids

3. PtCl4.6H2O can exist as a hydrated complex. 1 molal aqueous solution has depression in freezing point of 3.72º

Assume 100% ionisation and

then complex is

a. b.

c. d.

4. A weather balloon filled with hydrogen gas at 1 atm and 27ºC has volume equal to 12000 litres. On ascending, it reaches a place where temperature is –23ºC and pressure 0.5 atm. The volume of the balloon is

a. 24000 L b. 12000 L c. 10000 L d. 20000 L

5. In Bohr model of the hydrogen atom, the lowest orbit corresponds to

a. Infinite energy b. The maximum energy c. The minimum energy d. Zero energy

6. When 4KMnO is reduced with oxalic acid in acidic

solution, the oxidation number of M n changes from a. 7 to 4 b. 6 to 4 c. 7 to 2 d. 4 to 2

7. The unit of molar conductivity is

a. –1cm–2 mol–1 b. cm–2 mol–1

c. –1cm2 mol–1

d. cm2 mol–1

8. In a double bond connecting two atoms, there is a sharing of a. 2 electrons b. 1 electron c. 4 electrons d. All electrons

9. In a reversible reaction, the catalyst a. Increases the activation energy of the backward reaction b. Increases the activation energy of the forward reaction

c. Decreases the activation energy of both, forward and backward reaction

d. Decreases the activation energy of forward reaction

10. Ammonium cyanide is salt of NH4OH(Kb = 1.8 × 10–5)

and HCN (Kb = 4.0 × 10–10). The hydrolysis constant of

0.1 M at 25ºC is

a. 1.4 b.

c. d.

11. For the reaction A + B C it is found that doubling the

concentration of A increases the rate by 4 times, and

doubling the concentration of B doubles the reaction rate.

What is the overall order of the reaction?

a. 4 b. 3/2

c. 3 d. 1

12. The coagulation power of an electrolyte for arsenious

sulphide decreases in the order

a. + +3 +2Na , Al , Ba b. 3 24 4PO , SO , Cl- - -

c. +3 +2 +Al , Ba , Na d. None of these

13. Correct relationship between heat of fusion heat

of vaporisation and heat of sublimation is

a.

b.

c.

d.

14. The number of neutrons in the parent nucleus which gives

N14 on - emission is

a. 7 b. 14 c. 6 d. 8

15. Which represents nucleophilic aromatic substitution

reaction?

a. Reaction of benzene with in sunlight

b. Benzyl bromide hydrolysis

c. Reaction of NaOH with dinitrofluorobenzene

d. Sulphonation of benzene

16. Which one of the following is the chiral molecule?

a. b.

c. d.

17. n-Propyl chloride and benzene react in the presence of

anhydrous to form

a. ethyl benzene b. methyl benzene

c. n-propyl benzene d. iso-propyl benzene

1f 2K (H O) 1.86 mol kg,

2 6 4[Pt(H O) ]Cl 2 4 2 2 2[Pt(H O) Cl ]Cl 2H O

2 3 3 2[Pt(H O) Cl ],Cl.3H O 2 4 4 2[Pt(H O )Cl ]4H O

4NH CN

157.2 1017.2 10 61.4 10

fus( H ),

vap( H ) sub( H )

fus vap subH H H

vap fus subH H H

sub vap fusH H H

sub vap fusH H H

2Cl

3CH Cl 2 2CH Cl

3CHBr CHClBrI

3AlCl

Chemistry104

18. 1-chlorobutane reacts with alcoholic KOH to form

a. 1-butene b. 2-butane

c. 1-butanol d. 2-butanol

19. An alcohol having molecular formula C5H11OH on

dehydration gives an alkene, which on oxidation yield a

mixture of ketone and an acid. The alcohol is

a. b.

c. d.

20. m-chlorobenzaldehyde on reaction with conc. KOH at

room temperature gives

a. potassium m-chlorobenzate and m-chlorobenzyl alcohol

b. m-hydroxy benzaldehyde and m-chlorobenzyle alcohol

c. m-chlorobenzyl alcohol and m-hydroxy benzyle alcohol

d. potassium m-chlorobenzoate and m-hydroxy bezaldehyde

21. Acetamide is treated separately with the following

reagents. Which one of these would give methyl amine?

a. PCl5 b.

c. Sodalime d. Hot conc.

22. On treating aniline with nitrous acid and HCl at 0–5ºC

gives

a. An alcohol

b. Diazonium salt

c. Nitro aniline

d. Aniline hydrogen chloride

23. ‘Rayon’ is

a. Natural silk b. Artificial silk

c. Natural plastic or rubber d. Synthetic plastic

24. The waxes are long chain compounds of fatty acids, which

belong to the class of

a. Esters b. Ethers

c. Alcohols d. Acetic acid

25. Which of the following is a local anaesthetic?

a. Diazepam b. Procaine

c. Mescaline d. None of these

26. The electronic configuration of the element which is just

above the element with atomic number 43 in the same

periodic group is

a.

b.

c.

d.

27. Complex is formed in the extraction of

a. Na b. Cu c. Ag d. Fe

28. when heated gives

a. Magnesium oxychloride b. Magnesium dichloride

c. Magnesium oxide d. Magnesium chloride

29. Acidified potassium dichromate on reacting with a

sulphite is reduced to

a. 2 2CrO Cl b. 24CrO

c. 3Cr d. 2Cr

30. In the extraction of which of the following, complex ion

forms?

a. Cu b. Ag c. Fe d. Na

3 2 2 3CH CH CH(OH)CH CH 3 2 2 3CH CHCH CH CH|OH

3 2 3(CH ) CHCH(OH)CH 3 2 2(CH ) CCH OH

2NaOH Br

2 4H SO

2 2 6 2 6 5 21s 2s 2p 3s 3p 3d 4s

2 2 6 2 6 10 2 51s 2s 2p 3s 3p 3d 4s 4p

2 2 6 2 6 6 11s 2s 2p 3s 3p 3d 4s

2 2 6 2 6 10 1 61s 2s 2p 3s 3p 3d 4s 4p

2 2MgCl .6H O


105Mock Test-3

JEE ADVANCE PAPER-I


The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).

1. The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity

of a solution of a weak acid HY (0.10 M). If 0 0

X Y,

the difference in their apK values, apK (HX) apK (HY),

is (consider degree of ionization of both acids to be << 1)

2. The total number of and particles emitted in the

nuclear reaction is

3. The number of hydroxyl group(s) in Q is

4. The total number of basic groups in the following form of lysine is

5. In 1 L saturated solution of AgCl [Ksp(AgCl)=1.6×10–10],

0.1 mol of CuCl is added. The

resultant concentration of in the solution is

The value of “x” is

6. EDTA4– is ethylenediaminetetraacetate ion. The total number of N – Co – O bond angles in [Co(EDTA)1–] complex ion is

7. Reaction of Br2 with Na2CO3 in aqueous solution gives sodium bromide and sodium bromated with evolution of CO2 gas. The number of sodium bromide molecules involved in the balanced chemical equation is

8. A decapeptide (Mol. Wt. 796) on complete hydrolysis gives glycine (Mol. Wt.75), alanine and phenylanine. Glycine contributes 47.0% to the total weight of the hydrolysed products. The number of glycine units present in the decapeptide is



9. If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with m fraction of octahedral holes

occupied by aluminium ions and n fraction of tetrahedral holes occupied by magnesium ions, m and n, respectively, are

a. 1 1

,2 8

b. 1

1,4

c. 1 1

,2 2

d. 1 1

,4 8

10. A monatomic ideal gas undergoes a process in which the ratio of P to V at any instant is constant and equals to 1. What is the molar heat capacity of the gas?

a. b. c. d. 0

11. The given graph represents the variation of Z

(compressibility factor ) versus P, for three real

gases A, B and C. Identify the only incorrect statement.

a. For the gas A, a = 0 and its dependence on P is linear at all pressure

b. For the gas B, b = 0 and its dependence on P is linear at all pressure

c. For the gas C, which is typical real gas for which

neither a nor By knowing the minima and the

point of intersection, with Z = 1, a and b can be calculated

d. At high pressure, the slope is positive for all real gases

12. For the reaction: 3 2 4I ClO H SO 4 2Cl HSO I

The correct statement(s) in the balanced equation is/are

a. Stoichiometric coefficient of is 6

b. Iodide is oxidised c. Sulphur is reduced d. H2O is one of the products.

13. The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation of such a

reaction will be and

a. b.

c. d.

238 21492 82U Pb

6sp[K (CuCl) 1.0 10 ]

Ag

x1.6 10 .

4R

2

3R

2

5R

2

PV

nRT

b 0.

4HSO

1(R 8.314 JK log 2 0.301)153.6 kJ mol 148.6 kJ mol

158.6 kJ mol 160.5 kJ mol

1

A

B

C

P (atm)

Z

0

AIdeal gasBC

3 2 2 2 2H N CH CH CH CH

CH C

O

2H N O

4aqueous dilute KMnO (excess)Hheat 0 C

P Q

3H C

H

HO 3CH

Chemistry106

14. The freezing point (in ºC) of a solution containing 0.1 g of

K3[Fe(CN)6](Mol. Wt. 329) in 100 g of water (Kf = 1.86K

kg mol–1

) is

a. b.

c. d.

15. The carboxyl functional group (—COOH) is present in

a. picric acid b. barbituric acid

c. ascorbic acid d. aspirin

16. Which compound would give 5-keto-2-methyl hexanal

upon ozonolysis?

a. b.

c.

d.

17. For the process at T=100ºC and 1

atmosphere pressure, the correct choice is

a.

b.

c.

d.

18. Which of the following exists as covalent crystals in the

solid state?

a. Iodine b. Silicon

c. Sulphur d. Phosphorus


You will have to match entries in Column I with the entries in

Column II.

19. According to Bohr’s theory,

Total energy Kinetic energy

Potential energy Radius of nth orbit

Match the following:

Column I Column II

(A) Vn/Kn = ? 1. 0

(B) If radius of nth orbit

2. –1

(C) Angular momentum in

lowest orbital

3. –2

(D) 4. 1

a. A → 3 ; B → 2 ; C → 1 ; D → 4

b. A → 1 ; B → 2 ; C → 3 ; D → 4

c. A → 4 ; B → 2 ; C → 1 ; D → 3

d. A → 3 ; B → 4 ; C → 1 ; D → 2

20. An aqueous solution of X is added slowly to an aqueous

solution of Y as shown in Column - I. The variation in

conductivity of these reactions in Column - II.

Column I Column II

(A) 1. Conductivity

decreases and

then increases

(B) 2. Conductivity

decreases and

then does not

change much

(C) 3. Conductivity

increases and

then does not

change much

(D) 4. Conductivity

does not

change much

and then

increases

a. A → 1 ; B → 2 ; C → 3 ; D → 4

b. A → 2 ; B → 4 ; C → 3 ; D → 1

c. A → 3 ; B → 4 ; C → 2 ; D → 1

d. A → 3 ; B → 4 ; D → 1 ; D → 2

22.3 10

−− ×

25.7 10

−− ×

35.7 10

−− ×

21.2 10

−− ×

2 2H O( ) H O(g)→l

system surroundingS 0 and S 0∆ > ∆ >

system surroundingS 0 and S 0∆ > ∆ <

system surroundingS 0 and S 0∆ < ∆ >

system surroundingS 0 and S 0∆ < ∆ <

nE = nK =

nV =

nr =

x

nE , x ?∝ =

y

n

1Z , y ?

r∝ =

2 5 3 3X Y

(C H ) N CH COOH+

3X Y

KI(0.1M) AgNO (0.01 M)+

3X Y

CH COOH KOH+

X YNaOH HI+

3CH

3H C

3CH

3CH

3CH

3CH

3CH

3CH


107Mock Test-3




1. Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is

2. When the following aldohexose exists in its D–configuration, the total number of stereoisomers in its pyranose form is

3. The number of resonance structures for N is

4. The concentration of R in the reaction RP was measured as a function of time and the following data is obtained:

[R] (molar) 1.0 0.75 0.40 0.10

t (min). 0.0 0.05 0.12 0.18

The order of the reaction is

5. The oxidation number of Mn in the product of alkaline oxidative fusion of MnO2 is

6. The total number of diprotic acids among the following is

7. The number of neutrons emitted when undergoes

controlled nuclear fission to and is

8. In dilute aqueous 2 4H SO , the complex diaquodioxa-

latoferrate (II) is oxidised by 4MnO . For this reaction,

the ratio of the rate of change of [H ] to the rate of

change of 4[MnO ] is



9. The correct statement(s) for cubic close packed (cep) three dimensional structure is (are)

a. The number of the nearest neighbours of an atom

present in the topmost layer is 12

b. The efficiency of atom packing is 74%

c. The number of octahedral and tetrahedral voids per

atom are 1 and 2, respectively

d. The unit cell edge length is 2 2 times the radius of the

atom

10. The standard Gibbs energy change at 300 K for the

reaction is 2494.2 J. At a given time, the

composition of the reaction mixture is and

The reaction proceeds in the:

[ ]

a. forward direction because

b. reverse direction because

c. forward direction because

d. reverse direction because

11. A compound MpXq has cubic close packing (ccp)

arrangement of X. Its unit cell structure is shown below.

The empirical formula of the compound is

a. MX b.

c. d.

12. Which of the following is the energy of a possible excited

state of hydrogen?

a. b.

c. d.

13. Assuming 2s-2p mixing is not operative, the paramagnetic

species among the following is

a. Be2 b. B2

c. C2 d. N2+

14. In allene(C3H4), the type(s) of hybridization of the carbon

atoms is (are)

a. b.

c. d.

KCN 2 4K SO 4 2 2 4(NH ) C O

NaCl 3 2Zn(NO ) 3FeCl

2 3K CO 4 3NH NO LiCN

2

2

CHO|

CH|

CHOH|

CHOH|

CHOH|

CH OH

3 4H PO 2 4H SO 3 3H PO 2 3H CO 2 2 7H S O

3 3H BO 3 2H PO 2 4H CrO 2 3H SO23592 U

14254 Xe 90

38Sr

2A B C 12[A] , [B] 2

12[C] .

R 8.314 J / K / mol, e 2718

cQ K

cQ K

cQ K

cQ K

2MX

2M X 5 14M X

13.6 eV 6.8 eV

3.4 eV 6.8 eV

3sp and sp 2sp and sp

2only sp 2 3sp and sp

M =X =

Chemistry108

15. Isomers of hexane, based on their branching, can be divided into three distinct classes as shown in the figure

I.

II.

II.

The correct order of their boiling point is a. I > II > III b. III > II > I c. II > III > I d. III > I > II

16. (isomeric products)

C5H11Cl (isomeric products)

What are N and M ? a. 6, 6 b. 6, 4 c. 4, 4 d. 3, 3


Each paragraph describes an experiment, a situation or a problem. Two multiple choice questions will be asked based on this paragraph. One or more than one option can be correct.

Paragraph for Question Nos. 17 to 18 X and Y are two volatile liquids with molar weights of 10g mol–1 and 40g mol–1 respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.

17. The value of d in cm (shown in the figure), as estimated

from Graham’s law, is a. 8 b. 12 c. 16 d. 20

18. The experimental value of d is found to be smaller than the estimate obtained using Graham’s law. This is due to

a. larger mean free path for X as compared to that of Y b. larger mean free path for Y as compared to that of X c. increased collision frequency of Y with the inert gas as

compared to that of X with the inert gas d. increased collision frequency of X with the inert gas as

compared to that of Y with the inert gas


When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7ºC was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of

a strong acid with a strong base is a constant 1(–57.0 kJ mol ),

this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M

acetic acid 5a(K 2.0 10 ) was mixed with 100 mL of 1.0 M

NaOH (under identical conditions to Expt. 1) where a

temperature rise of 5.6 C was measured. (Consider heat

capacity of all solutions as 1 14.2 Jg K and density of all

solutions as 1.0 1g mL )

19. Enthalpy of dissociation (in kJ 1mol ) of acetic acid

obtained from the Expt. 2 is a. 1.0 b. 10.0 c. 24.5 d. 51.4

20. The pH of the solution after Expt. 2 is a. 2.8 b. 4.7 c. 5.0 d. 7.0

2Cl , h Nv

fractional distillation M

L = 24 cm

Cotton wool

Soaked in X

Cotton wool

Soaked in Y

Initial formation of

The product

d

CH3

CH3

H3C

and

and


109Mock Test-3

ANSWER & SOLUTIONS

JEE-Main

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

d a c d c c c c c b

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

c c c d b,c d c a c a

21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

b d b a b a c c c b

1. (d) 2 g of hydrogen occupy volume 22.4 L=

1 g of hydrogen occupies volume 22.4 1

11.2 L2

×= =

2. (a) ZnS has zinc blende type structure (i.e., ccp structure).

The S2–

ions are present at the corners of the cube and at the

centre of each face. Zinc ions occupy half of the tetrahedral

sites. Each zinc ion is surrounded by four sulphide ions

which are disposed towards the corner of regular tetrahedral.

Similarly, S2–

ions surrounded by four Zn2+

ions.

3. (c)

Hence,α = 1,∴ n = 2

Two species will be produced from single species which

is only possible for Cl3H2O

4. (d) 1

P 1atm,= 1

V 12000 L,=

1T 27 273 300K= + =

2

P 0.5 atm,= 2

V ?=

2T 23 273 250K= − + =

1 1 2 2

1 2

P V P V

T T=

or 1 1 2

2

2 1

P V T 1 12000 250V 20, 000 L

P T 0.5 300

× ×= = =

×

5. (c) In hydrogen atom, the lowest orbit (n = 1) corresponds

to minimum energy (– 13.6 eV).

6. (c) 5

In this reaction oxidation state of Mn change from +7 to +2.

7. (c) Molar conductivity

So, its unit will be Ω–1

cm2

mol–1

8. (c) In a double bond connecting two atoms sharing of 4

electrons take place as in H2C = CH2.

9. (c) Decreases the activation energy of both forward and

backward reaction.

10. (b) Wh

a b

KK

K K=

×

4

4 5

1 101.4

4.0 10 1.8 10

−

− −

×= =

× × ×

11. (c) A + B → C On doubling the concentration of A rate of

reaction increases by four times. Rate ∝[A]2. However on

doubling the concentration of B, rate of reaction increases

two times. Rate ∝[B]

Thus, overall order of reaction = 2 + 1 = 3

12. (c) According to Hardy-Schulze rule.

13. (c) Heats of combustion are always exothermic except

oxidation of N as,

;

;

14. (d)

in no. of neutrons 14 – 6 = 8.

15. (b, c)

16. (d)

A carbon atom which is attached to four different atoms

or groups is called a chiral or asymmetric carbon atom.

such a carbon atom is often marked by an asterisk.

17. (c)

18. (a)

19. (c)

f cal f( T ) K m∆ = ×

f cal( T ) 1.86 1 1.86∆ = × =

f obs

f cal

( T ) 3.72i 2 1 (n 1) .

( T ) 1.86α

∆= = = = + −

∆

∴

2 3 3 2[Pt(H O) Cl ],Cl3H O

COOH|COOH

27

4 2 4 2 4 4 2 22KMnO 3H SO K SO 2MnSO 10CO 8H O++

+ + → + + +

4 2 4 2 4 4 2 22KMnO 3H SO K SO 2MnSO 10CO 8H O+ + → + + +

1

Mρ=

12 2 22

N O N O+ → H ve∆ = +

2 2N O 2NO+ → H ve∆ = +

14 14

6 6 1X Nβ

+→

14

6X

*

H|

I — C — Br|Cl

3 2 2 2CH CH CH CH Cl KOH(alc.)− + →

3 2 2 21-butene

CH CH CH CH KCl H O− = + +

3 3

3

CH — CH — CH — CH||OHCH

Dehydration→

2 2 3CH CH CH

3AlCl

3 3 2CH CH CH Cl+ →

OH

Na

−

+→

DinitrofluorobenzeneNO

NO F

NO

2O OH

+

DinitrophenolDinitro phenol Dinitrofluorobenzene

Chemistry110

Oxidation

3O C — CH

|OH

20. (a)

21. (b)

“Hofmann’s bromamide reaction”

22. (d)

23. (b) ‘Rayon’ is man-made fibre which consists of purified cellulose in the form of long threads. Rayon resembles silk in appearance. Hence called as artificial silk.

24. (a) Waxes are esters of higher fatty acids.

25. (b) The anaesthetics produce temporary insensitibility to the vital function of all type of cells, specially of nervous system and are used during surgical operations.

These are classified as (a) General anasthetic – produces unconsciousness all over

the body e.g. Cyclopropane, chloroform

(b) Local anasthetic – affect only the part of body e.g. Xylocaine, Procain etc.

26. (a) .

27. (c) Hydrometallurgy

2 2 2Ag S 4NaCN 2Na[Ag(CN) ] Na S

2 2 42Na[Ag(CN) ] Zn Na [Zn(CN) ] 2Ag

28. (c)

29. (c) 2 2 3 22 7 3 4 2Cr O 8H 2SO 2Cr 3SO 4H O

30. (b) 2 22

Sodium dicyno argentate

Ag S 4NaCN 2Na Ag CN Na S

22Na[Ag(CN) ] Zn 2 4

Sodium tetracynozincate (ppt)

Na [Zn (CN) ] 2Ag

JEE Advance Paper-I

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

3 8 4 2 7 8 5 6 a a

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

b a, b, d a a d b b b a c

1. (3) HX H X

[H ][X ]Ka

[HX]

HY H Y

[H ][Y ]Ka

[HY]

m for 1mHX m for

2mHY

1 2m m

1

10

2Ka C

1

2

m1 1 0

m1

Ka C

2

2

m2 2 0

m2

Ka C

2

1 1 1

2 2 2

Ka C m

Ka C m

20.01 1

0.0010.1 10

1 2pKa pKa 3

2. (8)

total 8 particles.

3. (4)

3 3

3

CH — C CH — CH|

CH

3

3

CH — C O|

CH

O O|| |

Ph — C — H HO Ph — C — H|

OHI

3 2 2 3 2

O||

CH — C — NH Br NaOH CH NH

2N O,

5 225 Mn 3d 4s

heat2 2 2MgCl .6H O MgO 5H O 2HCl

238 6 214 2 21492 80 82U X Pb

(6 ,2 ),

H

+

H

HO

HOHO

OH

OH

(Q)

(P)4aqueous dilute KMnO

(excess) 0 C

NH2 2N Cl

NH2

O O

HClHONO

Cl

CHOCannizzaro's

reaction

KOH

Cl

COOK

Cl

2CH OH

ææÆ

111Mock Test-3

4. (2)

5. (7) Let the solubility of AgCl is x

and that of CuCl is y

. . .(i)

Similarly

. . .(ii)

On solving (i) and (ii)

6. (8) Total no. of N – Co – O bond angles is 8.

7. (5)

8. (6) Let number of glycine units Mass of decaeptide = 796

Mass of needed = 162 g, Total mass = 958 g

9. (a) In ccp lattice:

Number of O atoms

Number of Octahedral voids

Number of tetrahedral voids

Number of

Number of

Due to charge neutrality

10. (a)

11. (b)

12. (a, b, d)

13. (a)

14. (a) ,

fi

15. (d)

16. (b)

17. (b) At and 1 atmosphere pressure

is at equilibrium. For equilibrium

and

18. (b) Silicon exists as covalent crystal in the solid state.

19. (a) A 3; B 2; C 1; D 4

20. (c) A 3; B 4; C 2; D 1

(A)

Initially conductivity increases due to ion formation after that it becomes practically constant because X alone cannot form ions. Hence (3) is the correct match.

(B)

Number of ions in the solution remains constant until all the AgNO3 precipitated as AgI. Thereafter conductance increases due to increase in number of ions. Hence (4) is the correct match.

1mol litre

xxAgCl Ag Cl 1mol litre

y yCuCl Cu Cl

1spK of AgCl [Ag ] [Cl ]

101.6 10 x(x y)

spK of CuCl [Cu ] [Cl ] 61.6 10 y(x y)

7[Ag ] 1.6 10

x 7

22 3 3 23Br 3CO 5Br BrO 3CO

n

2H O

47958 75 n

100

958 47

n 6100 75

4

4

83Al 4 m

2Mg 8 n

4( 2) 4m( 3) 8n( 2) 0

1 1m and n

2 8

3 2 4ClO 6I 6H SO

2 4 23I Cl 6HSO 3H O

2 1

1 2

T TEa0.3010

2.303R T T

3

Ea 310 3000.3010

2.303 8.314 10 310 300

aE 53.6 kJ mol

33 6 6K [Fe(CN) ] 3K [Fe(CN) ] i 4

f f

m 1000 0.1 1000T K i 1.86 4

M W 329 100

22.3 10 2fT 2.3 10

100 C

2 2H O( ) H O(g)

totalS 0 system surroundingS S 0

system surroundingS 0 and S 0

2 5 3 3 2 5 3 3X Y

(C H ) N CH COOH (C H ) NH CH COO

3 3X Y

KI(0.1 M) A gNO (0.01 M) AgI KNO

3CH

3CH

3CH

3CH

CHOO

3

2

OZn / H O

5-keto-2-methyl hexanal

3

O||

O — C — CH

COOH

(Aspirin)

O

Co N O

O

N O

O O

O

O

and 2— NH are basic groups in lysine.

O||

—C — O

Chemistry112

(C) Initially conductance decreases due to the decrease in the

number of OH ions thereafter it slowly increases due to

the increase in number of H+ ions. Hence (2) is the correct

match.

(D) Initially it decreases due to decrease in H+ ions and then

increases due to the increase in OH ions. Hence (1) is the

correct match.


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

3 8 9 5 6 6 3 9 b,c,d b

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

b c c b b b c d a b

1. (3) LiCN are basic in nature and their

aqueous solution turns red litmus paper blue.

2. (8)

Total No. of stereoisomers = 24

= 16 which contains 8D –

Configuration and 8 – L Configuration.

3. (9)

4. (5) From two data, (for zero order kinetics)

⇒

5. (6)

6. (6) 3 3 2 3

H PO , H SO

7. (3)

8. (9) 0

0

a2.303K log ,

t a x

Ê ˆ= Á ˜-Ë ¯

9. (b, c, d) (a) For any atom in topmost layer, coordination

number is not 12 since there is no layer above topmost

layer, (b) Fact, (c) Fact, (D) 2 a 4R=

So, a 2 2 R=

10. (b)

i.e. backward reaction.

11. (b)

12. (c) Energy in 1st excited state

13. (c) Assuming that no 2s-2p mixing takes place

(a)

(b)

(diamagnetic)

(c)

(d)

14. (b)

KCN, 2 3K CO ,-

I

x 0.25K 5

t 0.05= = =

II

x 0.60K 5

t 0.12= = =

2 2 2 4 2 2 4(Potassiummanganate)

2MnO 4KOH O 2K MnO 2H O O.S. of Mn 6in K MnO+ + → + = +

2 2 2 4 2 2 42MnO 4KOH O 2K MnO 2H O O.S. of Mn 6in K MnO+ + → + = +

2 4 2 3 2 2 7 2 4 2 3H SO , H CO , H S O , H CrO , H3PO3, H SO

235 142 90 1

92 54 38 0U Xe Sr 3 n→ + +

0

1/80

a2.303K log

1ta

8

=

e cG RT ln K∆ = −

e c2494.2 8.314 300 ln K= ×

⇒1

cK e−=

1

c

1K e 0.36

2.718

−= = =

12

2 2

2(B)(C)Q 4

[A] [1/ 2]

×= = =

cQ K ,>

1 1X :8 6 4

8 2× + × =

1M 4 1 2;

4= × + =

2 4 2M X MX=

3.4 eV= −

2 2 2 22Be 1s , *1s , 2s , * 2s (diamagnetic)σ σ σ σ→

0x0y

2p2 2 2 2 22 z 2p

B 1s , *1s , 2s , *2s , 2p , (diamagnetic)π

πσ σ σ σ σ→

1 0x x1 0y y

2p *2p2 2 2 2 2 02 z z2p *2p

C 1s , *1s , 2s , *2s , 2p , , , *2p (paramagnetic)π π

π πσ σ σ σ σ σ→

1 0x x1 0y y

2p *2p2 2 2 2 2 02 z z2p *2p

C 1s , *1s , 2s , * 2s , 2p , , , *2p (paramagnetic)π π

π πσ σ σ σ σ σ

2 2 2 2 2 02 z zN 1s , *1s , 2s , * 2s , 2p , , , *2p (diamagnetic)

π π

π πσ σ σ σ σ σ→

2 0x x2 0y y

2p *2p2 2 2 2 2 02 z z2p *2p

N 1s , *1s , 2s , * 2s , 2p , , , * 2p (diamagnetic)π π

π πσ σ σ σ σ σ

2 2sp sp sp

H

HC C C== ==

H

H(allene)

–O O O

1 2 3

OO

7 8 9

–O

O O

4 5 6

–O

–O

OH NaOH N→

HO NaOH

N→

N is

H

H

HH HO

O OH

OH

OH2CH

113Mock Test-3

15. (b) III > II > I More the branching in an alkane, lesser will be the surface area, lesser will be the boiling point.

16. (b)

Md, 1 cannot be separated by fractional distillation.


17. (c)

18. (d) As the collision frequency increases then molecular speed decreases than that expected.


19. (a) 2HCl NaOH NaCl H O

n 100 1 100 m mole 0.1 mole

Energy evolved due to neutralization of HCl and NaOH

0.1 57 5.7 kJ 5700 Joule

Energy used to increase temperature of solution

200 4.2 5.7 4788 Joule Energy used to increase temperature of calorimeter

5700 4788 912 Joule ms. t 912

m.s 5.7 912 ms 160 Joule / C [Calorimeter constant]

Energy evolved by neutralization of 3CH COOH and

NaOH 200 4.2 5.6 160 5.6 5600 Joule So energy used in dissociation of 0.1 mole

3CH COOH 5700 5600 100 Joule Enthalpy of dissociation 1 kJ / mole

20. (b) 3

1 100 1CH COOH

200 2

3

1 100 1CH CONa

200 2

a

[salt]pH pK log

[acid]

1/ 2

pH 5 log2 log1/ 2

pH= 4.7

X

Y

r d 402

r 24 d 10

d 48 2d 3d 48d 16 cm

CH3

CH3

H3C Cl

CH3

CH3 H3C

*

1,d

CH3

CH3 H3C

Cl *

1,d

CH3

CH2Cl H3C

Chemistry114


1. The number of moles of 2 2SO Cl in 13.5 g is

a. 0.1 b. 0.2

c. 0.3 d. 0.4

2. Arrangement of sulphide ions in zinc blende is

a. simple cubic b. hcp

c. bcc d. fcc

3. A pressure cooker reduces cooking is increased

a. Heat is more evenly distributed

b. Boiling point of water inside the cooker is increased

c. The high pressure tenderizes the food

d. All of the above

4. Two flasks of equal volume contains 2SO and 2CO

respectively at 25 C and 2 atm pressure. Which of the

following is equal in them ?

a. masses of the two gas b. number of molecules

c. rates of effusion d. molecular structure

5. The ratio of the kinetic energy to the total energy of an

electron in a Bohr orbit is

a. –1 b. 2

c. 1 : 2 d. None of these

6. Oxygen has oxidation states of +2 in the

a. 2 2H O b. 2CO c. 2H O d. 2OF

7. Given The equivalent

conductance of the electrolytic cell is

a. b.

c. d.

8. In molecule, the atoms are bonded by

a. One , Two b. One , One

c. Two , One d. Three bonds

9. For the reaction the

equilibrium constant changes with

a. Total pressure

b. Catalyst

c. The amounts of and taken

d. Temperature

10. A certain weak acid has a dissociation constant of

The equilibrium constant for its reaction with a

strong base is.

a. b.

c. d.

11. According to Arrhenius theory, the activation energy is a. The energy it should possess so that it can enter into an

effective collision b. The energy which the molecule should possess in order

to undergo reaction c. The energy it has to acquire further so that it can enter

into a effective collison d. The energy gained by the molecules on colliding with

another molecule

12. Which of the following is property of colloid? a. Scattering of light b. They show attraction c. Dialysis d. Emulsion

13. Which of the following is an example of endothermic reaction?

a.

b.

c.

d.

14. The nuclear binding energy for (39.962384 amu) is: (given mass of proton and neutron are 1.007825 amu and 1.008665 amu respectively)

a. b.

c. d. None of these

15. Which is an electrophile?

a. b. c. d.

16. Cyanide and isocyanide are isomers of type a. Positional b. Functional c. Tautomer d. Structural

17. Acetone will be formed by the ozonolysis of a. Butene-1 b. Butene-2 c. Isobutene d. Butyne-2

18. Which of the following reactions gives

?

a. 3Zn /CH OHææææÆ

b.

c.

d.

1/ a 0.5 cm , R 50 ohm, N 1.0. l

1 2 110 ohm cm gm eq 1 2 120 ohm cm gm eq

1 2 1300 ohm cm gm eq 1 2 1100 ohm cm gm eq

2N

2 2H (g) I (g) 2HI(g),

2H 2I

41.0 10 .

41.0 10 101.0 10101.0 10 141.0 10

2 2 2 2 6C H 2H C H ; E 314.0kJ

2 2C O CO ; E 393.5kJ

2 2N O 2NO; E 180.5kJ

2 2 22H O 2H O; E 571.8kJ

Ar

343.81 MeV 0.369096 MeV

931 MeV

3AlCl CN 3NH 3CH OH

2 2H C C C CH

2 2CH Br CBr CH

2 3

o

Aq.K CO2

40 CHC C CH COOH

Zn2 2

HeatCH Br C C CH Br

2 22CH CH CH I

115Mock Test-4

19. Which of the following does not form phenol or phenoxide?

a. 6 5C H Cl b. 6 5C H COOH

c. 6 5 2C H N Cl d. 6 5 3C H SO Na 20. Identify the final product (Z) in the following sequence of

reactions:

a. b.

c. d.

21. When propionic acid is treated with aqueous sodium

bicarbonate, is liberated. The C of comes from?

a. methyl group b. carboxylic acid group c. methylene group d. bicarbonate group

22. Aniline undergoes condensation to form Schiff base on reacting with

a. Acetyl chloride b. Ammonia c. Acetone d. Benzaldehyde

23. The mass average molecular mass and number average molecular mass of a polymer are respectively 40,000 and 30,000. The polydispersity index of polymer will be

a. < 1 b. > 1 c. 1 d. 0

24. Hardening of oils is caused by

a. b. c. d.

25. Which of the following is molecular disease?

a. Allergy b. Cancer

c. German measeles d. Sickel-cell-anaemia

26. Hydrogen can be put in halogen group because

a. It has deuterium and tritium as isotopes

b. It forms hydrides like chlorides

c. It contains one electron only

d. It is light

27. Which of the following metal is extracted by

amalgamation process?

a. Tin b. Silver

c. Copper d. Zinc

28. Which of the following hydroxide is insoluble in water?

a. b.

c. d.

29. The product of oxidation of I ion by 4MnO in alkaline

medium is

a. 2I b. 3IO

c. 4IO d. 3I

30. Potassium ferrocyanide is a

a. Normal salt b. Mixed salt

c. Double salt d. Complex salt

3 2 4H O H SO2 Heat

Me C O HCN X Y Z;

3 2(CH ) C(OH)COOH 2 3CH C(CH )COOH

2 3HOCH CH(CH )COOH 3CH CH CHCOOH

2CO 2CO

2H 2N 2O 2CO

2Be(OH) 2Mg(OH)

2Ca(OH) 2Ba(OH)


Chemistry116

JEE ADVANCE PAPER-I



1. HCl gas is passed into water, yielding a solution of

density and containing 30% HCl by weight.

Calculate the molarity of the solution.

2. A sample contains a mixture of and

HCl is added to 15.0 g of the sample, yielding 11.0 g of

NaCl. What percent of the sample is ?

Reactions are:

Mw of Mw of Mw of

3. Amongst the following, the total number of compounds

soluble in aqueous is

4. The total number of contributing structures showing hyperconjugation (involving C—H bonds) for the following carbocation is.

5. The total number(s) of stable conformers with non-zero dipole moment for the following compound is (are)

6.

;

Total number of isomers (including stereoisomers) is 7. Total number of isomers

8. In the scheme given below, the total number of intramolecular aldol condensation products formed from ‘Y’ is



9. For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is Assuming concentration of solute is Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take

a. 724 b. 740 c. 736 d. 718 10. Two closed bulbs of equal volume (V) containing an ideal

gas initially at pressure pi and temperature T1 are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then I raised to T2. The final pressure pf is:

11.095g mL

3NaHCO 2 3Na CO .

2 3Na CO

2 3 2 2

3 2 2

Na CO 2HCl 2NaCl CO H O

NaHCO HCl NaCl CO H O

NaCl 58.5, 3NaHCO 84,1

2 3Na CO 106g mol

NaOH

3 2 2 2CH — CH — CH — CH OH

3 2 3

enantiomeric ( )

CH — CH — CH — CH|

OH

3 3

3

OH|

CH — C — CH|CH

3 2

3

CH — CH — CH OH|

CH

2 C.

1bK 0.76K kg mol )

1T

iP , V

1T

iP , V

1T

fP ,V

2T

fP ,V

3

2

1. O 1. NaOH(aq)2. Zn, H O 2. heat

Y

3CH 3CH

H H

meso

3CH

3CHH

H

3CH

3CHH

H

Pair of enantiomers

Mirror

3CH 2 3CH CH 3CH 3CH

Br

Cl

3CH

Cl

3CH

Br

2 3OCH CH

2CH OH

OH

2 3CH CH

2 3CH CH

COOH

2NO

N3H C

3CH

OHN3H C 3CH COOH

117Mock Test-4

a. b.

c. d.

11. A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron,

respectively, then the value of h/ (where is wavelength associated with electron wave) is given by:

a. meV b. 2 meV

c. d.

12. Reduction of the metal centre in aqueous permanganate ion involves

a. 3 electrons in neutral medium b. 5 electrons in neutral medium c. 3 electrons in alkaline medium d. 5 electrons in acidic medium

13. For the following electrochemical cell at 298 K,

Given:

The value of x is a. –2 b. –1 c. 1 d. 2

14. Which one of the following molecules is expected diamagnetic behaviour?

a. C2 b. N2

c. O2 d. S2

15. Aqueous solutions of HNO3, KOH, CH3COOH and CH3COONa of identical concentrations are provided. The pair (s) of solutions which form a buffer upon mixing is (are)

a. and

b. KOH and

c. and

d. and

16. The value of log10K for a reaction A B is

(Given

and 2.303 × 8.314 × 298 = 5705 )

a. 5 b. 10 c. 95 d. 100

17. The number of structural isomers for C6H14 is a. 3 b. 4 c. 5 d. 6

18. Amongst the given options, the compound(s) in which all

the atoms are in one plane in all the possible

conformations (if any), is (are)

a. b.

c. d.

SECTION 3 Contains 2 Matches The Following Type Questions


Column II.

19. According to Bohr’s theory,

Total energy Kinetic energy

Potential energy Radius of nth orbit

Match the following:

Column I Column II

(A) 1. 0

(B) If radius on nth orbit

2. –1

(C) Angular momentum in

lowest orbital

3. –2

(D) 4. 1

a. A 3; B 2; C 4; D 1

b. A 1; B 2; C 3; D 4

c. A 3; B 1; C 2; D 4

d. A 3; B 2; C 1; D 4

20. Match the thermodynamic processes given under Column

I with the expression given under Column II:

Column I Column II

(A) Freezing of water at

273 K and 1 atm

1. q = 0

(B) Expansion of 1 mol of

an ideal gas into a

vacuum under

isolated conditions

2. w = 0

(C) Mixing of equal

volumes of two ideal

gases at constant

3.

1 2i

1 2

T Tp

T T

1i

1 2

T2p

T T

2i

1 2

T2p

T T

1 2i

1 2

T T2p

T T

meV 2 meV

4 22Pt(s) | H (g, 1 bar) H (aq, 1 M) || M (aq), M (aq) | Pt(s)

2

cell 4

[M (aq)]E 0.092 V when 10 .

[M (aq)]x

4 2

0

M / M

RTE 0.151 V; 2.303 0.059 V

F

3HNO 3CH COOH

3CH COONa

3HNO 3CH COONa

3CH COOH 3CH COONa

o 1r 298KH 54.07 kJ mol , o 1 1

r 298KS 10 JK mol 1 1R 8.314 JK mol ;

2H C C O 2 2H C C CH

nE nK

nV nr

n nV / K ?

xnE , x ?

yn

1Z , y ?

r

sysS 0

2CHH C C — C

HH H

2H C 2CH

C C

Chemistry118

temperature and

pressure in an isolated

container

(D) Reversible heating of

at 1 atm from

300 K to 600 K,

followed by reversible

4.

cooling to 300 K at 1

atm

5.

a. A 3,5; B 1,2,4 ; C 1,2,4; D 1,2,4,5

b. A 1,2,4; B 1,2,4,5; C 3,4; D 1,2,4

c. A 3,5; B 1,2,4; C 1,2,4,5; D 1,2,4

d. A 1,2,4; B 3,5; C 1, 4,5; D 1,2,4

2H (g)

U 0 G 0


119Mock Test-4




9 (both inclusive).

1. 0.45 g of an acid (mol wt. = 90) required 20 ml of 0.5 N

for complete neutralization. Basicity of acid is

2. The co-ordination number of copper in cuprammonium

sulphate is

3. The co-ordination number of cobalt in the complex

is

4. The primary valence of the metal ion in the co-ordination

compoun is

5. The oxidation number of Cr in is

6. The number of equivalent Cr O bonds in 24CrO is.

7. The number of the following reagents that produce ppt.

with 4ZnSO solution is.

2 3 2 4 2 3 3NaOH, N CO , NaCl, Na HPO , Na S, CH CO Na

8. The change in the magnetic moment value when

2

2 4Cu H O

is converted to 2

3 4Cu NH

is.



9. A gas described by van der Waal’s equation

a. behaves similar to an ideal gas in the limit of large

molar volumes

b. behaves similar to an ideal gas in the limit of large

pressures

c. is characterized by van der Waal’s coefficients that are

dependent on the identity of the gas but are independent of

the temperature

d. has the pressure that is lower than the pressure exerted

by the same gas behaving ideally

10. Assuming that Hund’s rule is violated, the bond order and

magnetic nature of the diatomic molecule is

a. 1 and diamagnetic b. 0 and diamagnetic

c. 1 and paramagnetic d. 0 and paramagnetic

11. The kinetic energy of an electron in the second Bohr orbit

of a hydrogen atom is [a0 is Bohr radius].

a. b.

c. d.

12. The equilibrium in aqueous

medium at 25ºC shifts towards the left in the presence of

a. b. c. d.

13. The bond energy (in kcal mol–1) of C – C single bond is

approximately

a. 1 b. 10 c. 100 d. 1000

14. Which is correct statement if

is added at equilibrium condition?

a. The equilibrium will shift to forward direction because

according to IInd law of thermodynamics the entropy must

increases in the direction of spontaneous reaction

b. The condition for equilibrium is

where G is Gibbs free energy per mole of the gaseous

species measured at that partial pressure. The condition of

equilibrium is unaffected by the use of catalyst, which

increases the rate of both the forward and backward

reactions to the same extent

c. The catalyst will increase the rate of forward reaction

by α and that of backward reaction by .

d. Catalyst will not alter the rate of either of the reaction

15. For a first order reaction AP, the temperature (T)

dependent rate constant (k) was found to follow the

equation log k The pre-exponential

factor A and the activation energy Ea, respectively, are

a. and

b. and

c. and

d. and

16. The initial rate of hydrolysis of methyl acetate (1 M) by a

weak acid (HA, 1M) is 1/100th of that of a strong acid

(HX, 1M), at 25ºC. The Ka of HA is

a. b.

c. d.

KOH

2 2 2[Co(en) Br ]Cl

2 4K [Ni CN ]

3 6 3[Cr(NH ) ]Cl

2B

2

2 20

h

4 ma

2

2 20

h

16 ma2

2 20

h

32 ma

2

2 20

h

64 ma

I II2Cu Cu Cu

3NO Cl SCN CN

2 2 3N 3H 2NH 2N

2 2 3N H NHG 3G 2G

1(2000) 6.0.

T

6 11.0 10 s 19.2 kJmol

16.0s 116.6 kJmol

6 11.0 10 s 116.6 kJmol

6 11.0 10 s 138.3kJmol

41 10 51 1061 10 31 10

Chemistry120






Thermal decomposition of gaseous 2

X to gaseous X at 298 K

takes place according to the following equation:

2X (g) 2X(g)

The standard reaction Gibbs energy, rG∆ ° of this reaction is

positive. At the start of the reaction, there is one mole of 2

X

and no X. As the reaction proceeds, the number of moles of X

formed is given by .β Thus

equilibriumβ is the number of moles of

X formed at equilibrium. The reaction is carried out at a

constant total pressure of 2 bar. Consider the gases to behave

ideally. (Given: R = 0.083 L bar 1 1K mol )− −

17. The equilibrium constant pK for this reaction at 298 K, in

terms of equilibrium ,β is

a.

2

equilibrium

equilibrium

8

2

β

β− b.

2

equilibrium

2

equilibrium

8

4

β

β−

c.

2

equilibrium

equilibrium

4

2

β

β− d.

2

equilibrium

2

equilibrium

4

4

β

β−

18. The incorrect statement among the following, for this

reaction, is

a. Decrease in the total pressure will result in formation of

more moles of gaseous X

b. At the start of the reaction, dissociation of gaseous 2

X

takes place spontaneously

c. equlibrium 0.7β =

d. C

K 1<


Tollen’s reagent is used for the detection of aldehyde when a

solution of is added to glucose with then

gluconic acid is formed

Gluconic acid

[Use 2.303 and at

19.

Find ln K of this reaction.

a. 66.13 b. 58.38

c. 28.30 d. 46.29

20. When ammonia is added to the solution, pH is raised to

11. Which half-cell reaction is affected by pH and by how

much?

a. will increase by a factor of 0.65 from

b. will decrease by a factor of 0.65 from

c. will increase by a factor of 0.65 from

d. will decrease by a factor of 0.65 from

3AgNO

4NH OH

Ag e Ag ;+ −+ → o

redE 0.8 V=

6 12 6 2C H O H O+ →

6 12 7(C H O ) 2H 2e ;+ −+ + o

oxdE 0.05 V=−

3 2 3Ag(NH ) e Ag(s) 2NH ;+ −+ → + o

oxdE 0.337 V=

RT0.0592

F× =

F38.92

RT= 298 K

6 12 6 2 6 12 72Ag C H O H O 2Ag (s) C H O 2H− ++ + → + +

oxdE

o

oxdE

oxdE

o

oxdE

redEo

redE

redE

o

redE


121Mock Test-4

ANSWER & SOLUTIONS

JEE-Main

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

a d b b a d a a d c

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

c a a a a b c c b c

21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

d d b a b b b a b d

1. (a) Molecular mass of

2 2SO Cl 32 2 16 35.5 2 135= + × + × =

Moles 13.5

0.1135

= =

2. (d) Arrangement of sulphide ions 2

(S )−

in zinc blende

(ZnS) is fcc while 2Zn +

ions occupy alternate tetrahedral

voids.

3. (b) The temperature at which a liquid boils increases with

increase in pressure.

4. (b) Equal volumes of all gases under similar conditions of

temperature and pressure contain equal number of

molecules.

5. (a) K.E. = – (T.E.)

6. (d) Oxygen have oxidation state in

7. (a)

8. (a)

9. (d) Equilibrium constant changes with temperature,

pressure and the concentration of either reactant or

product.

10. (c)

Clearly, the reverse reaction is the hydrolysis reaction.

11. (c) The definition of activation energy.

12. (a) Scattering of light is a property of colloid.

13. (a) For exothermic reactions

For endothermic reactions

14. (a) Total no of protons

Total no of neutrons

Mass defect

Binding energy = mass defect 931

15. (a) is lewis acid i.e., electron deficient compound.

So it is electrophile.

16. (b) and are functional isomers.

17. (c)

18. (c)

19. (b) Benzoic acid.

20. (b)

21. (d)

22. (d)

23. (b) Average number molecular weight

Average mass molecular weight

Polydispersity index (PDI)

24. (a)

25. (b) “Cancer” is known as molecular disease.

2+ 2OF .

1/ a 0.5 cm , R 50 ohm−= =l

Ra 50p 100

0.5= = =

l

1 2 11000 1 1000 1 1000k 10 ohm cm gm eq

N p N 100 1

− −Λ = × = × = × =

4

aHA : K 10−=

HA NaOH NaA H O+ + 2HA NaOH NaA H O+ +

⇒4

10aRequired 14

h w

K1 10K 10

K K 10

−

−= = = =

p RH H .<

p RH H .>

40

18Ar 18=

22=

[m p m n] 39.962384= × + × −

[1.007825 18 1.008665 22] 39.962384= × + × −

[18.14085 22.19063] 39.962384= + − 0.369=

×

0.369 931 343.62MeV= × =

3AlCl

CyanideR — C N≡≡

IsocyanideR — N C=

r

3 2

3

CH — C CH|CH

==

3

2

O

H O, Zn→

3

3Acetone

CH C O|CH

==

Zn

2 2CH Br C C CH Br∆

− ≡ − →2 2

CH C C CH= = =

2Me C O HCN= + →3

3(X)

OH|

CH — C — CN|

CH

3H O+

→

3

3(Y)

OH|

CH — C — COOH|

CH

2 4H SO→ 2

3(Z)

CH CCOOH|

CH

==

* *

3 2 3 3 2 2 2CH — CH — COOH Na HCO CH CH COONa H O CO+ → + +

* *

3 2 3 3 2 2 2CH — CH — COOH Na H CO CH CH COONa H O CO+ → + +

6 5 2 6 5 6 5 6 5 2Schiff 's base

C H NH O CHC H C H N CHC H H O+ = → − = +

nM 30,000=

wM 40,000=

w

n

M 40,0001.33

30,000M= = =

Ni

2Oil(unsaturated) H Fat (saturated)+ →

N N

π σ

π

Chemistry122

26. (b) Hydrogen, forms hydrides like halides, e.g. HCl.

27. (b) 2 2 2 2Cu Cl Ag S Cu S 2AgCl

2 22AgCl Hg Hg Cl 2Ag

AgCl Hg Ag HgCl

28. (a) The solubility of hydroxides of alkaline earth metals in water increases on moving down the group.

29. (b) 24 3KI MnO K IO Mn

30. (d) In 4 6K Fe(CN) , the species retains its identity in solid

as well as in solution state.


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

9 9 4 6 3 5 7 1 a c

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

d a, d d c c. d b c b, c d a

1. (9)

2. (9) Let x be the percentage of Then, weight of

Moles of NaCl produced

The NaCl is produced by the reaction of mol of

and mol of Each mol of

produces 2 mol of NaCl.

Solve x :

3. (4) Aromatic alcohols and carboxylic acids form salt with NaOH, will dissolve in aqueous NaOH.

4. (6) These are total to carbon and they all can

participate in hyperconjugation.

5. (3)

6. (5)

7. (7)

8. (1)

9. (a)

10. (c) Initial moles = final moles

11. (d)

12. (a, d) In acidic medium

In neutral medium,

Hence, number of electrons loose in acidic and neutral medium are 5 and 3 respectively.

13. (d)

2

% by weight 10 30 10 1.0959 M

36.5

dM

Mw

2 3Na CO .

3NaHCO (15 )g x

11.0g0.18mol

58.5g

106

x

2 3Na CO(15 )

84

x3NaHCO .

2 3Na CO

2 15

0.188106 84

x x2 313.5g Na CO ,

3NaHCO (15 1.35) 13.6g

2 3 2 3

1.35% Na CO 100 9.0% Na CO

15

6 — H 2sp

L bT k m sb

s solution

Wk

M W

s

2.5 10002 0.76

M 100

s

0.76 2.5 1000M 9.5

100 2

ss

solution

n760 xX

760 n

s

s

solv

solv

W

W 2.5 1836.0

W 9.5 100M

x 760 36 724

i i f f

1 1 2 1

P V P V P V P V

RT RT RT RT

i i f f

1 1 2 1

P P P P

T T T T i

f1 2 1

2P 1 1P

T T T

i 1 2f

1 1 2

2P T TP

T T T

2

f i1 2

TP 2P

T T

K.E. eV

h

2meV h

2meV

24 2MnO 8H 5e Mn 4H O

4 2 2MnO 2H O 3e MnO 4OH

2 2o

cell cell 10 42

0.059 [M ][H ]E E log

2 [M ]pH

x10

0.0590.092 0.151 log 10

2

x 2

H

HH

Three structures Two structures

2 3CH CH3H C

OHCOOH COOH

3H C3CH

OH

N

123Mock Test-4

14. (c) O2 is expected to be diamagnetic in nature but actually

it is paramagnetic.

15. (c, d) In option (c), if HNO3 is present in limiting amount

then this mixture will be a buffer and the mixture given in

option (d), contains a weak acid (CH3COOH) and its salt

with strong base NaOH, i.e. CH3COONa.

16. (b)

Hence (b) is correct.

17. (c)

, Hence (c) is correct

18. (b, c)

19. (d) A → 3; B → 2; C →1; D →4

20. (a) A 3,5; B 1,2,4; C 1,2,4; D 1,2,4,4


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

2 4 6 2 3 4 4 0 a, c, d a

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

c b,c,d c b d a b c b a

1. (2)

2. (4) In Cuprammonium sulphate co-

ordination no. of is 4.

3. (6)

number of bidentate ligand

number of monodentate ligand

4. (2) Primary valencies are also known as oxidation state.

5. (3)

x – 3 = 0, x + 3, Oxidation number of Cr is = +3

6. (4)

7. (4)

8. (0)

9. (a, c, d)

At low pressure, when the sample occupies a large

volume, the molecules are so far apart for most of the time

that the intermolecular forces play no significant role, and

the gas behaves virtually perfectly.

a and b are characteristic of a gas and are independent of

temperature. The term represents the pressure

exerted by an ideal gas while P represents the pressure

exerted by a real gas.

10. (a)

Bond order (nature is diamagnetic as no

unpaired electron)

11. (c) and ⇒

⇒

⇒ . . .(i)

Expression for

⇒ . . .(ii)

⇒ . . .(iii)

For n = 2 K.E.

12. (c, d) Cu2+

ions will react with CN–

and SCN–

forming

and leading the reaction in

the backward direction.

G H T S 54.07 1000 298 10∆ ° = ∆ ° − ∆ ° = − × − ×1

1057050 J mol 57050 5705log K−= − − = −

10log K 10=

6 14C H

3 2 2 2 2 3H C — CH — CH — CH — CH — CH

3 2 2 3

3

H C — CH — CH — CH — CH|CH

3 2 3

3

H C — CH — CH — CH — CH|

CH

3 3

33

H C — CH — CH — CH||CHCH

3

3 2 3

3

CH|

H C — C — CH — CH|CH

→ → → →

BW 1000

Normality = N =Eq.wt V

×

×

∴0.45 1000

Eq. Wt 450.5 20

×= =

×

∴Molec. Wt 90

Basicity 2Eq. Wt 45

= = =

3 4 4[Cu(NH ) ]SO

Cu

2 2 2[Co(en) Br ]Cl

C.N.of Co 2= ×

1+ × 2 2 1 2 6.= × + × =

2 4K [ Ni(CN) ], 2 x 4 0 x 2+ − = ⇒ = +

x 6 (0) 3 ( 1) 0+ × + × − =

2

2

n aP (V nb) nRT

V

+ − =

2

2

n aP

V

+

2 2 2 2 2x

* *

2 1s 1s 2s 2s 2pB (10) σ σ σ σ π=

6 41

2

−= =

nhmvr

2π=

2 2

2

mv e

r r= 2 2mv r e=

2e 2v

nh

π×= ∴

nh(mvr )

2π=

4 2

2

2 2

1me 4

1 2K.E. mv2 n h

π×= =

2

0 2 2

ha

4 meπ=

22

2

0

hme

4 aπ=

2

2 2 2

0

h 1K.E.

8ma nπ= ×

2

2 2

0

h

32 maπ=

3

4[Cu(CN) ] − 3

4[Cu(SCN) ] −

Chemistry124

also combines with which reacts with Cu to

produce CuCl pushing the reaction in the backward direction.

13. (c)

14. (b)

15. (d) Given, log

Since, log

So,

and Ea

16. (a) Rate in weak acid (rate in strong acid)


17. (b) Paragraph-1 2X (g) 2X(g)

Initial mole 1 0

eq.t t (1 ) 2

Given equilibrium2 So, equilibrium

2

Total mole at equilibrium eq(1 ) (12

eq.

eq eq2 total total total

eq. eq eq

1 2 22Px P P P2 2

12

eq eq

x(g) total totaleq eq

2P P P

21

2

So

2

eq.total2

eq.

p2 eq.

totaleq

2P

2(Px)K

(Px ) 2P

(2 )

2 2eq. eq

p total2 2eq. eq

4 8K P

4 4

18. (c) (a) Correct statement. As one decrease in pressure reaction will move in the direction where no. of gaseous molecules increases.

(b) Correct statement

At the start of reaction p pQ K so dissociation of 2X take

place spontaneously. (c) Incorrect statement as

2 2eq

p 2 2eq

8 8 (0.7)K

4 4 (0.7)

1, but

(d) Correct statement.

As pG 0 & G RT n K

pG 1, so K should be less than 1.

So K 1

ng.p cK K (RT) (RT 1)

p

c

KK

RT c pK K

So cK 1


19. (b)

20. (a) On increasing concentration of NH3 the concentration of H+

ion decreases. Therefore, Ered increases.

22Cu 2CN Cu(CN)

2 22Cu(CN) 2CuCN (CN) 3

4CuCN 3CN [Cu(CN) ] 2 3

4Cu 4SCN [Cu(SCN) ] 2Cu

2CuCl

2CuCl Cu 2CuCl

2000K 6

T

K log EaA

2.303RT

6 1A 10 sec38.3kJ / mole

1

100

weak acid strong acid

1[H ] [H ]

100

2weak acid

1[H ] M 10 M

100

2C 10

4aK 10

ocell

RTE ln K

nF

1 0.0592(0.8 0.05) ln K

2 2.303

(0.8 0.05) 22.303ln k 58.38

0.0592

125Mock Test-5


1. One mole of potassium dichromate completely oxidises the following number of moles of ferrous sulphate in acidic medium

a. 1 b. 3 c. 5 d. 6

2. In a solid lattice, the cation has left a lattice site and is located at an interstitial position, the lattice defect is

a. Frenkel defect b. Schottky defect c. F-centre defect d. Valency defect

3. At 300 K, the vapour pressure of an ideal solution containing 3 mole of A and 2 mole of B is 600 torr. At the same temperature, if 1.5 mole of A and 0.5 mole of C (non-volatile) are added to this solution the vapour pressure of

solution increases by 30 torr. What is the value of

a. 940 b. 405 c. 90 d. None of these

4. If the root mean square speed of helium is 4.75 m s–1 at

25 C, then its speed will become 9.50 m s–1 at

a. 100 C b. 323 C c. 919 C d. 1192 C

5. Which electronic level would allow the hydrogen atom to absorb a photon but not to emit a photon?

a. 3s b. 2p

c. 2s d. 1s

6. The product of oxidation of I with 4M nO in alkaline

medium is

a. 3IO b. 2I c. IO d. 4IO

7. The standard reduction electrode potentials of four

elements are

and The element that displaces A from its

compounds is a. B b. C c. D d. None of these

8. The bond angle in carbon tetrachloride is approximately

a.

b.

c.

d.

9. The formation of nitric oxide by contact process

43.200 kcal is favoured by

a. Low temperature and low pressure b. Low temperature and high pressure c. High temperature and high pressure d. High temperature and excess reactants concentration

10. The pH of an aqueous solution is 9.0. If the

solubility product of is what is ?

a. b.

c. d. 0.1

11. If ‘I’ is the intensity of absorbed light and C is the concentration of AB for the photochemical process

the rate of formation of is directly

proportional to a. C b. I

c. d. C.I

12. Gold number is maximum for the lyophilic sol is a. Gelatin b. Haemoglobin c. Sodium oleate d. Potato starch

13. For the allotropic change represented by equation

the enthalpy change is

If 6 g of diamond and 6 g of graphite are

separately burnt to yield carbon dioxide, the heat liberated in the first case is

a. Less than in the second case by

b. More than in the second case by

c. Less than in the second case by

d. More than in the second case by

14. The half-life of if its K or is is

a. b.

c. d.

15. Most stable carbonium ion is

a. b.

c. d.

16. Which of the following compounds will exhibit cis-trans isomerism

a. 2-butene b. 2-butyne c. 2-butanol d. Butanone

oBP ?

A 0.250 V, B 0.136 V, C 0.126 V

D 0.402 V.

90

109

120

180

2 2N O 2NO. H

2Mg OH

2Mg OH 111 10 , 2[ ]Mg

51 10 41.0 1021 10

AB hv AB*, AB*

2I

C(diamond) C(graphite);

H 1.89 kJ.

1.89kJ

1.89kJ

11.34 kJ

0.945 kJ

146C 42.31 10

22 10 yrs 33 10 yrs43.5 10 yrs 34 10 yrs

2 5C H

3 3(CH ) C

6 5 3(C H ) C

6 5 2C H CH

Chemistry 126

17. Acetylene gas is obtained by the electrolysis of a. Sodium fumarate b. Sodium succinate c. Sodium maleate d. Both (a) and (c) 18. The compound added to prevent chloroform to form

phosgene gas is

a.

b.

c.

d.

19. Epoxides are a. Cyclic ethers b. Not ethers c. Aryl-alkyl ethers d. Ethers with another functional group

20. reacts with

a. b.

c. d.

21. In the reaction

The compound X is a. Phthalic anhydride b. Phthalic acid c. o-xylene d. Benzoic acid

22. In the reaction

; C is

a. Pentanal b. Pentanone c. 2-Hexanone d. Hexanal

23. A polymer containing nitrogen is a. Bakelite b. Dacron c. Rubber d. Nylon-66

24. The base present in DNA, but not in RNA is a. Guanine b. Adenine c. Uracil d. Thymine

25. Which of the following is an antidiabatic drug a. Insulin b. Penicillin c. Chloroquine d. Aspirin

26. The correct sequence of elements in decreasing order of first ionisation energy is

a. b.

c. d.

27. In the metallurgical extraction of zinc from ZnO the reducing agent used is

a. Carbon monoxide b. Sulphur dioxide c. Carbon dioxide d. Nitric oxide

28. In the lime (kiln), the reaction

goes to completion because a. Of high temperature

b. CaO is more stable than

c. escapes simultaneously

d. CaO is not dissociated

29. Amalgams are a. Highly coloured alloys b. Always solid c. Alloys which contain mercury as one of the contents d. Alloys which have great resistance to abrasion

30. Generally, a group of atoms can function as a ligand if a. They are positively charged ions b. They are free radicals c. They are either neutral molecules or negatively charged ions d. None of these

2 5C H OH

3CH COOH

3 3CH COCH

3CH OH

NaOH / H

6 5 3C H OCH 3CH OH

3 3

O||

CH — C — CH 2 5C H OH

3NH8 6 4C H O X

Zn, HCl3 2 4CH (CH ) CN A

HONO OA B C

Na Mg Al Mg Na Al

Al Mg Na Mg Al Na

3 2CaCO (s) CO (g)

3CaCO

2CO


127Mock Test-5

JEE ADVANCE PAPER-I



1. The change in the number of unpaired electrons when

2

2 6Fe H O

is changed into 4

6Fe CN

is

2. A compound of mol. wt. 180 is acetylated to give a

compound of mol. wt. 390. The number of amino groups

in the initial compound is?

3. Specific rotations of -anomer of glucose is 112 and

for -anomer is +19 . Specific rotation of equilibrium

mixture is 52.6 . Calculate % composition of -and -

anomers in the equilibrium mixture.

4. Consider all possible isomeric ketones including

stereoisomers of MW = 100. All these isomers are

independently reacted with 4NaBH (Note: Stereoisomers

are also reacted separately). The total number of ketones

that give a racemic product(s) is/are

5. Amongst the following, the total number of compounds

soluble in aqueous NaOH is

6. The half-life period of a radioactive substance is 2 min.

The time taken for 1 g of the substance to reduce to 0.25 g

will be

7. of hypothetical MgCl is and for

is The enthalpy of

disproportionation of MgCl is –49x. Find the value of x.

8. Hydrolysis of an alkyl halide (RX) by dilute alkali

takes place simultaneously by and pathways. A

plot of vs. is a straight line of

the slope equal to and intercept equal to

Calculate the initial rate (mol of

consumption of RX when the reaction is carried out taking

of RX and 0.1 mol of ions.



9. The ionic radii (in Å) of and are

respectively:

a. 1.36, 1.40 and 1.71

b. 1.36, 1.71 and 1.40

c. 1.71, 1.40 and 1.36

d. 1.71, 1.36 and 1.40

10. From the following statements regarding choose

the incorrect statement

a. It can act only as an oxidizing agent

b. It decomposes on exposure to light

c. It has to be stored in plastic or wax lined glass bottles in

dark

d. It has to be kept away from dust

11. was added to an aqueous KCl solution

gradually and the conductivity of the solution was

measured. The plot of conductance versus the volume

of is

a.

b.

c.

d.

f H 1125kJ mol

2MgCl 1642 kJ mol .

[OH ]

2SN 1SN

1 d[R X]

[RX] dt

[OH ]

3 1 12 10 mol L h 2 11 10 h . 1 1L min )

11mol L 1L [OH ]

3 2N , O F

2 2H O ,

3AgNO (aq.)

(^)

3AgNO

volume

(d)

++

++

+ + +

volume

(c)

++ ++ ++++

volume

(b)

+

+ + +

++

++

+ ++

+

volume

(a)

OCH2CH3

CH2OH3

OH N

H3C CH3

NO2 OH

N

H3C CH3

CH2CH3 CH2OH3 COOH

COOH

Chemistry 128

12. The intermolecular interaction that is dependent on the

inverse cube of distance between the molecules is:

a. ion-ion interaction b. ion-dipole interaction

c. London force d. hydrogen bond

13. According to Molecular Orbital Theory

a. 22C is expected to be diamagnetic

b. 22O is expected to have a longer bond length than 2O

c. 2N and 2N have the same bond order

d. 2He has the same energy as two isolated He atoms

14. For the first order reaction

a. the concentration of the reactant decreases exponent-

tially with time

b. the half-life of the reaction decreases with increasing

temperature

c. the half-life of the reaction depends on the initial

concentration of the reactant

d. the reaction proceeds to 99.6% completion in eight

half-life duration

15. In the reaction, the time taken for 75%

reaction of P is twice the time taken for 50% reaction of P.

The concentration of Q varies with reaction time as shown

in the figure. The overall order of the reaction is

a. 2 b. 3 c. 0 d. 1

16. Choose the correct reason(s) for the stability of the

lyophobic colloidal particles.

a. Preferential adsorption of ions on their surface from the

solution

b. Preferential adsorption of solvent on their surface from

the solution

c. Attraction between different particles having opposite

charges on their surface

d. Potential difference between the fixed layer and the

diffused layer of opposite charges around the colloidal

particles

17. The standard enthalpies of formation of

and glucose(s) at are –400 kJ/mol, –300 kJ/mol and

–1300 kJ/mol, respectively. The standard enthalpy of

combustion per gram of glucose at is

a. + 2900 kJ b. – 2900 kJ c. – 16.11 kJ d. +16.11 kJ

18. Which of the following compounds is not colored yellow?

a.

b.

c.

d.



Column II.

19. Match each of the diatomic molecules in Column I with its property/properties in Column II.

Column - I Column – II

(A) 1. Paramagnetic

(B) 2. Undergoes oxidation

(C) 3. Undergoes reduction

(D) 4. Bond order ≥ 2

5. Mixing of ‘s’ and ‘p’ orbitals

a. A 5, 3, 1, B 5, 4, C 2, 1, D 4, 1, 2

b. A 1, 3, 5, B 1, 2, 4 C 3, 2, D 1, 2,

c. A 1, 3, B 1, 2, 4 C 1, 2, 3 D 4, 5,

d. A1, 3, 5, B4, 5, C1, 2, D1, 2, 4

[According to MOT]

20. Match the compounds/ions in Column I with their

properties/reactions in Column II.

Column I Column II

(A) 1. gives precipitate with 2, 4-dinitrophenyl-

hydrazine

(B) 2. gives precipitate with

(C) 3. is a nucleophile

(D) 4. is involved in cyano-hydrin formation

a. A 1, 2, 3, B 2, C 1, 3, 4, D 2, 3

b. A 1, 3, 4, B 2, 3 C 1, 2, 3, D 2

c. A 1, 3, B 1, 4, C 1, 3, D 2, 3

d. A 2, 3, B 1, 2, C 1, 3, D 1, 2

2 5 2 22N O (g) 4NO (g) O (g)

P Q R S

2CO (g), H2O( )l

25 C

25 C

2 6Zn [Fe(CN) ]

3 2 6K [Co(NO ) ]

4 3 3 10 4(NH ) [As(Mo O ) ]

4BaCrO

2B

2N

2O

2O

6 5C H CHO

3CH C CH3AgNO

CN

I

Time

[Q]

0[Q]

129Mock Test-5




9 (both inclusive).

1. The number of rings formed in is

……………

2. The number of equivalent bonds in is

……………

3. In fructose, the possible optical isomers are……………

4. Oxidation of glucose is one of the most important

reactions in a living cell. What is the number of ATP

molecules generated in cells from one molecule of

glucose?

5. 10 gm of a mixture of hexane and ethanol reacts with

sodium to give 200 ml. of 2H at 27 C and 760 mm

pressure. The percentage of ethanol is……………

6. The polymerisation of propene to linear polypene is

represented by the reaction

n

Where n has large integral value, the average enthalpies of

bond dissociation for and at 298 K

are +590 and respectively. The enthalpy

of polymerization is Find the value of n.

7. In the case of a first order reaction, the time required for

93.75% of reaction to take place is x times that required

for half of the reaction. Find the value of x.

8. In 1 L saturated solution of spAgCl (K of AgCl

0.1mol of CuCl (Ksp is

added. The resultant concentration of in the solution

is Calculate the value of x.



9. Which of the following atoms has the highest first

ionization energy?

a. Rb b. Na

c. K d. Sc

10. Among the electrolytes and

the most effective coagulating agent for

sol is

a. b. c. d.

11. The compound(s) with TWO lone pairs of electrons on the

central atom is(are)

a. 5BrF b. 3ClF

c. 4XeF d. 4SF

12. Solubility product constant of salts of types

and at temperature ‘T’ are

and respectively.

Solubilities (mole of the salts at temperature ‘T’

are in the order

a. b.

c. d.

13. According to the Arrhenius equation

a. a high activation energy usually implies a fast reaction b. rate constant increases with increase in temperature.

This is due to a greater number of collisions whose energy exceeds the activation energy

c. higher the magnitude of activation energy, stronger is

the temperature dependence of the rate constant d. the pre-exponential factor is a measure of the rate at

which collisions occur, irrespective of their energy

14. For a linear plot of log (x/m) versus log p in a Freundlich adsorption isotherm, which of the following statements is

correct?

a. Both k and 1/n appear in the slope term b. 1/n appears as the intercept

c. Only 1/n appears as the slope d. log (1/n) appears as the intercept.

15. When is adsorbed on a metallic surface, electron

transfer occurs from the metal to The TRUE

statement(s) regarding this adsorption is(are)

a. is physisorbed

b. heat is released

c. occupancy of of is increased

d. bond length of is increased

2Ca EDTA

Cl O 2 7Cl O

3

2

CH|CH CH

3

2 n

CH|

—CH — CH—

(C C) (C C)1331kJ mol ,

1360 kJ mol .

101.6 10 ), 6CuCl 1.0 10 )

Ag

x1.6 10 .

2 4Na SO , 2CaCl , 2 4 3Al (SO )

4NH Cl, 2 3Sb S

2 4Na SO 2CaCl 2 4 3Al (SO ) 4NH Cl

sp(K )

2MX, MX 3M X

8 144.0 10 , 3.2 10 152.7 10 ,3dm )

2 3MX MX M X 3 2M X MX MX

2 3MX M X MX 3 2MX M X MX

2O

2O .

2O

*2p 2O

2O

Chemistry 130

16. Methylene blue, from its aqueous solution, is adsorbed on

activated charcoal at For this process, the correct

statement is

a. The adsorption requires activation at

b. The adsorption is accompanied by a decrease in

enthalpy

c. The adsorption increases with increase of temperature

d. The adsorption is irreversible






Carbon–14 is used to determine the age of organic material.

The procedure is based on the formation of by neutron

capture in the upper atmosphere

is absorbed by living organisms during photosynthesis.

The content is constant in living organism once the plant or

animal dies, the uptake of carbon dioxide by it ceases and the

level of in the dead being, falls due to the decay which

undergoes

The half life period of is 5770 years. The decay constant

can be calculated by using the following formula

The comparison of the activity of the dead

matter with that of the carbon still in circulation enables

measurement of the period of the isolation of the material from

the living cycle. The method however, ceases to be accurate

over periods longer than 30,000 years. The proportion of

to in living matter is

17. Which of the following option is correct?

a. In living organisms, circulation of from atmosphere

is high so the carbon content is constant in organism

b. Carbon dating can be used to find out the age of earth

crust and rocks

c. Radioactive absorption due to cosmic radiation is equal to

the rate of radioactive decay, hence the carbon content

remains constant in living organism

d. Carbon dating cannot be used to determine concen-

tration of in dead beings

18. What should be the age of fossil for meaningful

determination of its age?

a. 6 years

b. 6000 years

c. 60,000 years

d. It can be used to calculate any age


Rocket propellants consist of rocket engines powered by

propellants. These are used both in space vehicles as well as in

offensive weapons such as missiles. The propellants are

chemical substances which on ignition provide thrust for the

rocket to move forward. These substances are called rocket

propellants. A propellant is a combination of an oxidiser and a

fuel which when ignited undergoes combustion to release large

quantities of hot gases. The passage of hot gases through the

nozzle of the rocket motor provides the necessary thrust for the

rocket to move forward according to Newton's third law of

motion.

19. A biliquid propellant contains

a. Liquid hydrazine

b. A mixture of liquid fuel and a liquid oxidiser

c. A solid rocket fuel

d. A liquid fuel which can also act as an oxidiser

20. A hybrid rocket propellant uses

a. A liquid oxidiser and a solid fuel

b. A composite solid propellant

c. A biliquid propellant

d. A solid, liquid and gas as a propellant

25 C.°

25 C°

14 C

14 1 14 1

7 0 6 1N n C n+ → +14 C

14 C

14 C

14C

14 14

6 7C N β −→ +14 C

( )λ

1/ 2

0.693.

tλ = -β

14 C

12 C121 : 10 .

14 C

14 C


131Mock Test-5

ANSWER & SOLUTIONS

JEE-Main

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

d a c c d a c b d d

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

b d d b c a d a a c

21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

a d d d a d a d c c

1. (d) 3

2 7Cr O Cr

6n

−− +→=

; Fe Fe1n

++ +++→=

eq. of 2 2 7

K Cr O = eq. of 4

FeSO 1 6 x 1× = ×

2. (a) When an ion (generally cation due to its small size) is

missing from its normal position and occpy an interstitial

site between the lattice points, the lattice defect obtained

is known as Frenkel defect.

3. (c)

and

4. (c) r.m.s. speed T∝

4.75 298

9.50 T= or

24.75 298

9.50 T

=

or T 1192 K= or T 1192 273 919 C= − = °

5. (d) When electron jumps to lower orbit photons are

emitted while photons are absorbed when electron jumps

to higher orbit, 1s orbital is the lower most, electron in

this orbital can absorb photons but cannot emit.

6. (a)

7. (c) A is displace from D because D have a

8. (b) is hybridised so bond angle will be

approximately

9. (d) High temperature and excess concentration of the

reactant concentration.

10. (d)

11. (b) In photochemical reaction the rate of formation of

product is directly proportional to the intensity of

absorbed light.

12. (d) Gold number shows the protective power of a

lyophilic solution. Lesser the gold number, greater will be

the protecting power of that colloid. Gelatin is one of the

best protective colloid. Among the given colloids, potato

starch has maximum gold number.

13. (d)

or

For combustion of by

14. (b)

15. (c) In the triphenyl methyl carbonium ion the electrons

of all the three benzene rings are delocalised with the

vacant p-orbital of central carbon atom. So, it is resonance

stabilised. It is the most stable of all the carbonium ions

given

The ion is stabilised by hyperconjugation, a

second order resonance.

16. (a)

Cis-trans isomerism shown by compound which have double

or triple bond by which they restrict their rotation, since 2

butyne have no hydrogen on triple bonded carbon.

[It does not show cis-trans]

17. (d)

Sodium salt of maleic acid or fumaric acid

18. (a) Because it float over chloroform and prevent its

oxidation.

19. (a) Cyclic ethers are called epoxies.

o o

A A B BP P x P x= +

o o o o

A B A B

3 2600 P P ;3P 2P 3000

3 2 2 3

= + + = + +

o o o o

A B A B

4.5 2630 P P ;4.5P 2P 4410

4.5 2 0.5 4.5 2 0.5

= + + = + + + + o o

A A1.5P 1410;P 940= = o

BP 90=

2

4 4 3 26MnO I 6OH 6MnO IO 3H O− − − − −+ + → + +

E 0.402 V° = −

4CCl 3sp

109 .°

pH 9= ∴ 9[H ] 10+ −=14

5

9

1 10[OH ] 10

10

−− −

−

×= =

2Mg(OH) Mg 2OH2Mg(OH) Mg 2OH+ −+

2 2

spK [Mg ][OH ]+ −= ⇒ 11 2 5 21 10 [Mg ][10 ]− + −× =

⇒11

2 1

5 2

1 10[Mg ] 10 0.1

(10 )

−+ −

−

×= = =

(graphite) (diamond)C C , H 1.9kJ→ ∆ =

(graphite) 2 2 1C O CO , H H+ → ∆ = −∆

(diamond) 2 2 2C O CO , H H+ → ∆ = −∆

1 2( H ) ( H ) 1.9 kJ− ∆ − − ∆ =

2 1H H 1.9∆ = ∆ +

2 16g, H H∆ > ∆ 1.9 / 2 0.95kJ.=

4

1/ 2 4

0.693 0.693t 0.3 10 yrs

k 2.31 10−= = = ×

×33.0 10 yrs.= ×

π

3

3

3

CH|

CH — C|CH

+

3 32 butyne

CH — C C — CH≡≡

CH — COONa||CH — COONa

→CHCOO||CHCOO

−

−2Na++

CHCOO||CHCOO

−

−→

Acetylene

CH|||CH

22CO 2e−+ +

2 2CH — CH

O

3CH

HC C==

3CH

HCis 2 butene

3CH

HC C==

3

H

CHTrans 2 butene

Chemistry 132

20. (c)

21. (a)

22. (d)

23. (d) Nylon-66- It is a polymer containing nitrogen

24. (d) Thymine is present in DNA while in RNA there is

Uracil. 25. (a) Insulin is an antidiabatic drug. 26. (d) This is due to the presence of fully

filled s-orbital in Mg. 27. (a)

2ZnO CO CO Zn

28. (d) Lime stone – CaCO3, Clay – silica and alumina Gypsum – CaSO4

29. (c) Amalgams are alloys which contain mercury as one of the contents.

30. (c) Ligand must have capacity to donate lone pair of electrons to form coordinate bond.

JEE Advance Paper -I 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

4 5 2 5 4 2 8 5 c a

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

d b a, c c b,c,d b c b b a

1. (4)

2. (5) Difference in mass of compound = 390 – 180 = 210

wt. of – group is = 43

Therefore no. of group = .

3. The -anomer of D-glucose has a specific rotation of +112

degrees in water. The β-anomer of D-glucose has a specific rotation of + 19 degrees (18.7 actually, but rounding up to 19).

If the fraction of glucose present as the -anomer is xand

the fraction present as the β-anomer is y, and the rotation of the mixture is +52.6 degree, we have X (+112.2 degree) + y(18.7) = 52.6 degree ……..(i) There is a very little of the open chain form present, so the fraction present as the α-anomer (x) plus the fraction present as the β-anomer (y) should account for all the glucose, i.e., x + y = 1 or y = 1 – x. Putting value of y in equation (i), we get x (+112.2 degree) + (1 – x)(18.7) = 52.6 degree ……..(ii) Solving equation (ii) for (x), we have, x = 0.36 or 36%. Thus (y) must be (1 – 0.36) = 0.64 or 64%.

So, percentage composition of - and -anomers in the

equilibrium mixture is 36% and 64% respectively.

4. (5)

5. (4)

6. (2)

7. (8)

;

;

8. (5) (by pathway)

rate constant of SN2 reaction

This is the equation of a straight line for

vs plot with slope equal to and intercept equal

to

dil NaOH3 32CH CO CH

3 2 3(Diacetone alchol)

3

OH O| ||

CH — C — CH — C — CH|CH

Zn , HCl3 2 4CH (CH ) CN

HONO3 2 4 2 2CH (CH ) CH NH

O3 2 4 2 3 2 4

Hexanal

CH (CH ) CH OH CH (CH ) CHO

2 6 2 4

nNylon-66

H H O| |||

—N — (CH ) — N — C — (CH ) — C —||O

Mg Al Na.

4 2.2H O

3CH CO

2NH210

4.88 543

1/ 2 1/ 2t t1g 0 5g 0 25g 25%t 4min

22MgCl Mg MgCl H ?

2Mg(s) 1/ 2Cl (g) MgCl 11H 125kJ mol

2 2Mg(s) Cl (g) MgCl 12H 642 kJ mol

12 1H H 2 H 642 (2 125) 392 kJ mol

49 392 x 8x

2

d[RX]k [RX][OH ]

dt

2SN

2k

2 1

d[RX]k [RX][OH ] k [RX]

dt

2 1

1 d[RX]k [OH ] k

[RX] dt

1 d[RX]

[RX] dt

[OH ]2k

1k .

N

H3C CH3

OH OH

N

H3C CH3

COOH

2CONH

COOH3NH

COOH

COOHC||O

O||C

133Mock Test-5

From question:

and

Hence,

9. (c) Ionic Radii order:

10. (a) It can act as an oxidising as well as reducing agent.

11. (d)

12. (b) Ion-dipole interaction

13. (a,c) (a) 22C Total no. of electrons = 14 so it is

diamagnetic

(b) 22O Bond order = 3; 2O Bond order = 2

Bond length in 22O is less than bond length in 2O .

(c) Bond order of 2N 2.5 ; Bond order of 2He 1/ 2

Some energy is released during the formation of 2He from

two isolated He atoms.

14. (a, b, d) For first order reaction

Hence concentration of decreases exponentially,

Also, , which is independent of concentration

and decreases with the increase of temperature.

15. (d) Overall order of reaction can be decided by the data

given

It is a first order reaction with respect to P.

From graph [Q] is linearly decreasing with time, i.e., order

of reaction with respect to Q is zero and the rate

expression is

16. (a, d) (a) Preferential adsorption of ions on surface from

the solution

(c) Attraction between particles having same charges on

their surface accounts for the Brownian motion.

(d) Definition of Zeta Potential.

17. (c) Combustion of glucose

18. (a) is bluish white ppt.

19. (d) A 1, 3, 5, B 4, 5, C 1, 2, D 1, 2, 4

[According to MOT]

20. (a) A 1, 2, 3, B 2, C 1, 3, 4, D 2, 3

(Note: Assuming is ammonical)

(a)

(b)

(c) ;

(d)

JEE Advance Paper –II

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

5 6 8 38 10 5 4 7 d c

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

b, c d b,c,d b,c,d b b c b 2 1

1. (5) 2. (6)

3. (8) Fructose has three chiral centres and hence

optical isomers are possible.

4. (38)

5. (10) 22400 ml of 2H is produced by 46 gms of ethanol.

200 ml of 2H is produced by 46 200 23

22400 56

3 1 1 2 12 1k 2 10 mol L hr , k 1 10 hr

[RX] 1.0M [OH ] 0.1M

3 2d[RX]2 10 1 0.1 1 10 1

dt

1 1300 mol L hr 1 15mol L min 3 2N O F

kt0[A] [A] e

2[NO ]

1/ 2

0.693t

K

1/ 2t

99.6

2.303 100t log

K 0.4

99.6 1/ 2

2.303 0.693t (2.4) 8 8 t

K K

75% 50%t 2t

1 0r k[P] [Q] .

6 12 6 2 2 2C H O 6O 6CO 6H O

combustion f 2 f 2 f 6 12 6H (6 H CO 6 H H O) H C H O

(6 400 6 300) ( 1300)

2900 kJ / mol

2900 /180 kJ / g

16.11 kJ / g

2 6Zn [Fe(CN) ]

3AgNO

3NH2

(whiteppt)

PhCHO Ag O PhCOO Ag

KCNPhCHO

CN|

Ph — C — O|H

3ammonical AgNO3 3

(White ppt)

CH C CH CH C C Ag

KCNPhCHO CN|

Ph — C — O|H

3AgNO CN AgCN

3AgNO I AgI

32 8

6 12 6 2 2 2C H O 6O 6CO 6H O 38ATP

2NH NH 2PhCHO O N

NO2

(ppt.)PhHC N NH

O2N

NO2

Volume

++

++

+ + +

Chemistry 134

Percentage of 2 5

23C H OH 100 9.6% 10%

56 10

6. (5) Energy released = Energy due to formation of two

single bonds of propene

polymerisation/mol

polymerisation

7. (4)

Also,

8. (7)

;

The value of x = 7

9. (d) I. P1 = Sc > Na > K > Rb

10. (c) As is a negative sol, so, will be the

most effective coagulant due to higher charge density on

in accordance with Hardy-Schulze rule. Order of effectiveness of cations:

11. (b, c)

One lone pair on central atom (Br) Two lone pair

on central atom (Cl)

Square planar

Two lone pair on central atom (Xe) One lone pair

on central atom (S)

Only, 3 4ClF & XeF contains two lone pair of electrons on

central atom.

12. (d) Solubility of

Solubility of

Solubility of

13. (b, c, d) Ea / RTK A e

2

dk EaK

dT RT

dkEa

dT

A Frequency factor

= No of collisions per unit time per unit volume.

14. (c) According to the Freundlich adsorption isotherm

15. (b, c, d) Adsorption of on metal surface is exothermic.

During electron transfer from metal to electron

occupies orbital of Due to electron transfer to the bond order of

decreases hence bond length increases.

16. (b) Adsorption of methylene blue on activated charcoal is physical adsorption hence it is characterised by decrease in enthalpy.

17. (c) Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay, hence the carbon content remains constant in living organism

18. (b) 6000 years

19. (b) Biliquid Propellant – A double base propellant is a high strength, high modulus gel of cellulose nitrate (gun cotton) in glyceryl trinitrate or a similar solvent.

20. (a) Hybrid Propellant – A hybrid propellant consists of a solid fuel and liquid oxidiser to provide propulsion energy and working substance e.g., Solid acrylic rubber and

liquid .

12 331 662 kJ mol

H 1590 662 72 kJ mol

H 72 n 360 n 5

1/ 21

0.0693t

k

93.751

2.303 100t log

k 100 93.75

1

2.303 100log

k 6.25

4

1

2.303log 2

k 1/ 2

1 1

4 2.303 log 2 4 0.6934t

k k

10sp

(S x)SAgCl Ag Cl (K 1.6 10 )

6sp

x (x S )CuCl Cu Cl (K 1 10 )

10S (S x) 1.6 10 6x(x S ) 10

4S1.6 10

x

4 4 6S (1.6 10 )x. x(x 1.6 10 x) 10

3x 10

7 xS 1.6 10 1.6 10

2 3Sb S 2 4 3Al (SO )

3Al

34Al Ca Na NH

8 4(MX) 4 10 2 10 5

2(MX ) 8 10 4

3(M X) 1 10

3 2MX M X MX .

1/ nxkP

m

x 1log log K log P

m n

2O

2O*2p 2O .

2O 2O

2 4N O

Xe

F

FF

FS

FF

(See-Saw Shape)

FF

Cl F

F

F

(T-Shape)

F

Br

F

F

F

F

(Square Shape)

135Mock Test-1

Mock Test “JEE-Main”





prohibited.






questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response.

6. Candidates will be awarded marks as stated above in instruction No. 5 for correct response of each question. 1/4 (one

fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will

be made if no response is indicated for an item in the answer sheet.

7. There is only one correct response for each question. Filling up more than one response in each question will be treated as

wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.

8. Use Blue/Black Ball Point Pen only for writing particulars/marking responses on Side-1 and Side-2 of the Answer Sheet.






11. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall.







Roll Number: in figures

in words




A Test Booklet code

Mathematics136

Read the Following Instructions Carefully:







5. For each incorrect response, one-fourth (¼) of the total marks allotted to the question would be deducted from

the total score. No deduction from the total score, however, will be made if no response is indicated for an item

in the Answer Sheet.











Invigilator on duty and sign the Attendance Sheet again. Cases where a candidate has not signed the

Attendance Sheet a second time will be deemed not to have handed over the Answer Sheet and dealt with as an

unfair means case. The candidates are also required to put their left hand THUMB impression in the

space provided in the Attendance Sheet.

12. Use of Electronic/Manual Calculator and any Electronic Item like mobile phone, pager etc. is prohibited.






hall/room.

137Mock Test-1

JEE-MAIN: MATHEMATICS MOCK TEST-1

1. The domain of definition of is:

a. b.

c. d.

2. How many roots the equation have

a. One b. Two

c. Infinite d. None

3. If is the cube root of unity, then +

a. 4 b. 0

c. – 4 d. None of these

4. From the following find the correct relation

a. b.

c. d.

5. If the roots of the cubic equation

are in G.P., then

a. b. c. d.

6.

a. b.

c. d.

7.

a. b.

c. d.

8. How many words can be made from the letters of the

word INSURANCE, if all vowels come together.

a. 18270 b. 17280


9. There are four machines and it is known that exactly two

of them are faulty. They are tested, one by one, in a

random order till both the faulty machines are identified.

Then, the probability that only two tests are needed, is

a. b.

c. d.

10.

a. 1/2 b. – ½ c. 1/4 d. 1

11. Find real part of

a. – 1 b. 1


12. From a 60 meter high tower angles of depression of the

top and bottom of a house are α and β respectively. If

the height of the house is 60sin ( )

,x

β α−then x=

a. sin sinα β b. cos cosα β

c. sin cosα β d. cos sinα β

13. The function is not defined

at The value which should be assigned to f at

so that it is continuous at is

a. b.

c. d.

14. If ,sin,cos 44 θθ ayax == then ,dy

dx at

3,

4

πθ = is

a. –1 b. 1

c. 2a− d.

2a

15. The minimum value of [(5 )(2 )]/[1 ]x x x+ + + for non-

negative real x is

a. 12 b. 1 c. 9 d. 8

16.

a. b.

c. d.

17. The measurement of the area bounded by the co-ordinate

axes and the curve is

a. 1 b. 2

c. 3 d.

18. The solution of the equation is

a. b.

c. d.

2

2

log ( 3)( )

3 2

xf x

x x

+=

+ +

1, 2R − − − ( 2, )− ∞

1, 2, 3R − − − − ( 3, ) 1, 2− ∞ − − −

2 21

1 1x

x x− = −

− −

ω 2 2(3 5 3 )ω ω+ +2 2(3 3 5 )ω ω+ + + =

( )AB A B′ ′ ′= ( )AB B A′ ′ ′=

1 adj AA

A

− = 1 1 1( )AB A B− − −=

3 2 0ax bx cx d+ + + =

3 3c a b d= 3 3ca bd= 3 3a b c d= 3 3ab cd=

1 2 3 42 3 4 ....

nC C C C nC+ + + + + =

2n . 2nn

1. 2nn − 1. 2nn +

2 4(log ) (log )

12! 4!

e en n+ + + =K

n 1/ n

11( )

2n n

−+1

( )2

n ne e

−+

1

3

1

6

1

2

1

4

3sin sin

10 10

π π =

1cosh (1)−

log(1 ) log(1 )( )

ax bxf x

x

+ − −=

0.x =

0x = 0,x =

a b− a b+

log loga b+ log loga b−

1tan

21

xedx

x

−

=+∫

2log(1 )x c+ +1tanlog xe c−

+1

tan xe c−

+1

1 tantan xe c−− +

logey x=

∞

2x y ydye x e

dx

− −= +

3

3

y x xe e c= + + 2y xe e x c= + +

3y xe e x c= + + xy e c= +

Mathematics138

19. The distance between and is

a. b. 4

c. d. None of these

20. The area of a circle whose centre is (h, k) and radius a is

a. 2 2 2( )h k aπ + − b. 2a hkπ

c. 2aπ d. None of these

21. The locus of the mid-point of the line segment joining the

focus to a moving point on the parabola is

another parabola with directrix

a. b.

c. d.

22. If and then the vector parallel to

is

a. (3, 5) b. (1, 1)

c. (1, 3) d. (8, 5)

23. The acute angle between the line joining the points (2, 1,

–3), (–3,1,7) and a line parallel to

through the point (–1, 0, 4) is

a. b.

c. d.

24. Which of the following is logically equivalent to

a. b.

c. d.

25. The number of solutions of the system of equations

is

a. 3 b. 2 c. 1 d. 0

26. The sum to infinity of the given series

is

a. b.

c. d.

27. The values of A and B such that the function

, is continuous

everywhere are

a. b.

c. d.

28.

a. b.

c. d. None of these

29. The equations of tangents to the circle

2 2 22 4 25 0x y x y+ − − + = which are perpendicular to the

line 5 12 8 0x y+ + = are

a. 12 5 8 0, 12 5 252x y x y− + = − =

b. 12 5 0, 12 5 252x y x y− = − =

c. 12 5 8 0,12 5 252 0x y x y− − = − + =

d. None of these

30. is logically equivalent to

a.

b.

c.

d.

4 3 11+ =x y 8 6 15,+ =x y

7

2

7

10

2 4y ax=

x a= −2

ax = −

0x =2

ax =

(2, 5)a =r

(1, 4),b =r

( )a b+r

r

1 3

3 4 5

x y z− += =

1 7cos

5 10

−

1 1cos

10

−

1 3cos

5 10

−

1 1cos

5 10

−

~ (~ )⇒p q

∧p q ~∧p q

~ ∧p q ~ ~∧p q

2 7, 3 2 1, 4 3 5x y z x y z x y z+ − = − + = + − =

2 3 4

1 1 1 1

2 3 4n n n n− + − +K

1loge

n

n

+

log1

e

n

n

+

1loge

n

n

−

log1

e

n

n

−

2sin ,2

( ) sin ,2 2

cos ,2

x x

f x A x B x

x x

π

π π

π

− ≤ −

= + − < <

≥

0, 1A B= = 1, 1A B= =

1, 1A B= − = 1, 0A B= − =

3 43 5x x dx+ =∫4 3/ 2(3 5 )x c+ + 4 3/ 21

(3 5 )5

x c+ +

4 3/ 21(3 5 )

30x c+ +

~ ∧p q

→p q

→q p

~ ( )→p q

~ ( )→q p


139Mock Test-1

JEE ADVANCE PAPER-I

Time 3 Hours. Max. Marks 264 (88 for Mathematics)








4. Section 3 contains 2 “match the following” type questions and you will have to match entries in Column I with the entries

in Column II.

Marking Scheme: For each entry in Column I, +2 for correct answer, 0 if not attempted and –1 in all other cases.


marking scheme too.






integer in the ORS.

Marking scheme:



1. If area enclosed between the curves and

and the axis of x is sq unit, then the value of

622

must be

2. The number of distinct solutions of the equation

2 4 4 6 65

cos 2 cos sin cos sin 24

+ + + + =x x x x x

in the

interval [0, 2 ]π is

3. Let the curve C be the mirror image of the parabola 2 4=y x with respect to the line 4 0.+ + =x y If A

and B are the points of intersection of C with the line

5,= −y then the distance between A and B is

4. The minimum number of times a fair coin needs to be

tossed, so that the probability of getting at least two heads

is at least 0.96 is

5. Let n be the number of ways in which 5 boys and 5 girls

can stand in a queue in such a way that all the girls stand

consecutively in the queue. Let m be the number of ways

in which 5 boys and 5 girls can stand in a queue in such a

way that exactly four girls stand consecutively in the

queue. Then the value of m

nis

6. TP and TQ are any two tangents to a parabola and the

tangent at a third point R cuts then in P' and Q', then the

value of must be

7. Let : →f R R be a function defined by

[ ], 2( ) ,

0, 2

x xf x

x

≤=>

where [x] is the greatest integer less

than or equal to x. If 2 2

1

( ),

2 ( 1)−

=+ +∫xf x

I dxf x

then the value

of (4 1)−I is

8. A cylindrical container is to be made from certain solid

material with the following constraints: It has a fixed

inner volume of 3mm ,V has a 2 mm thick solid wall and

is open at the top. The bottom of the container is a solid

circular disc of thickness 2 mm and is of radius equal to

the outer radius of the container.

ln( )y x e= +

1lnx

y

=

λ

TP TQ

TP TQ

′ ′=

Mathematics140

If the volume of the material used to make the container is

minimum when the inner radius of the container is 10

mm, then the value of / 250V π is


This section contains TEN questions.





Marking scheme:

+4 If only the bubble(s) corresponding to all the correct

option(s) is(are) darkened.



9. Let ∆PQR be a triangle. Let , ,= =uuur uuurrr

a QR b RP and

.=uuurr

c PQ If | | 12, | | 4 3= =rr

a b and . 24,=r rb c then which

of the following is (are) true?

a. 2| |

| | 122

− =r

rca b.

2| || | 30

2− =

rrca

c. | | 48 3× + × =rr r r

a b c a

d. . 72= −rr

a b

10. Let X and Y be two arbitrary, 3 3,× non-zero, skew-

symmetric matrices and Z be an arbitrary 3 3,× non-zero,

symmetric matrix. Then which of the following matrices

is (are) skew symmetric?

a. 3 4 4 3−Y Z Z Y b.

44 44+X Y

c. 4 3 3 4−X Z Z X d.

23 23+X Y

11. Which of the following values of α satisfy the equation

2 2 2

2 2 2

2 2 2

(1 α) (1 2α) (1 3α)

(2 α) (2 2α) (2 3α) 648α ?

(3 α) (3 2α) (3 3α)

+ + +

+ + + = −

+ + +

?

a. 4− b. 9

c. 9− d. 4

12. In 3 ,R consider the planes 1 : 0=P y and 2 : 1.+ =P x z

Let 3P be a plane, different from 1P and 2 ,P which passes

through the intersection of 1P and 2 .P If the distance of

the point (0, 1, 0) from 3P is 1 and the distance of a point

(α, β, γ) from 3P is 2, then which of the following

relations is (are) true?

a. 2α + β + 2γ 2 0+ =

b. 2α – β + 2γ 4 0+ =

c. 2α + β – 2γ – 10 0=

d. 2α – β + 2γ – 8 0=

13. In 3 ,R let L be a straight line passing through the origin.

Suppose that all the points on L are at a constant distance

from the two planes 1 : 2 1 0+ − + =P x y z and

2 : 2 1 0.− + − =P x y z Let M be the locus of the feet of the

perpendiculars drawn from the points on L to the plane

1 .P Which of the following points lie(s) on M?

a. 5 2

0, ,6 3

− −

b. 1 1 1

, ,6 3 6

− −

c. 5 1

,0,6 6

−

d. 1 2

,0,3 3

−

14. Let P and Q be distinct points on the parabola 22=y x

such that a circle with PQ as diameter passes through the

vertex O of the parabola. If P lies in the first quadrant and

the area of the triangle ∆OPQ is 3 2, then which of the

following is (are) the coordinates of P?

a. (4, 2 2)

b. (9,3 2)

c. 1 1

,4 2

d. (1, 2)

15. Let ( )y x be a solution of the differential equation

(1 ) 1.′+ + =x xe y ye If (0) 2,=y then which of the

following statements is (are) true ?

a. ( 4) 0− =y

b. ( 2) 0− =y

c. ( )y x has a critical point in the interval ( 1,0)−

d. ( )y x has no critical point in the interval ( 1,0)−

16. Consider the family of all circles whose centers lie on the

straight line y = x. If this family of circles is represented

by the differential equation 1 0,′′ ′+ + =Py Qy where P, Q

are functions of x, y and 2

2(here , ),′ ′ ′′= =

dy d yy y y

dx dx then

which of the following statements is (are) true?

a. = +P y x

b. = −P y x

c. 21 ( )′ ′+ = − + + +P Q x y y y

d. 2( )′ ′− = + − −P Q x y y y

141Mock Test-1

17. Let : →g R R be a differential function with

(0) 0, (0) 0′= =g g and (1) 0.′ ≠g

Let ( ), 0

( ) | |00,

≠= =

xg x x

f x xx

and | |( ) = xh x e for all

.∈x R Let ( )f ho (x) denote ( ( ))f h x and ( )( )h f xo

denote ( ( )).h f x Then which of the following is (are)

true?

a. f is differentiable at 0x =

b. h is differentiable at 0x =

c. f ho is differentiable at 0x =

d. h fo is differentiable at 0x =

18. Let ( ) sin sin sin6 2

=

π πf x x for all ∈x R and

( ) sin2

=π

g x x for all .∈x R Let (f o g)(x) denote f (g(x))

and (g o f )(x) denote g(f (x)). Then which of the following

is (are) true?

a. Range of f is 1 1

,2 2

−

b. Range of f o g is 1 1

,2 2

−

c. 0

( )lim

( ) 6→=π

x

f x

g x

d. There is an ∈x R such that ( )( ) 1g f x =o


This section contains TWO questions.

Each question contains two columns, Column I and Column II

Column I has four entries (A), (B), (C) and (D)

Column II has five entries (1), (2), (3), (4) and (5)

Match the entries in Column I with the entries in Column II

One or more entries in Column I may match with one or more

entries in Column II

The ORS contains a 4 5× matrix whose layout will be similar to

the one shown below:

(A) (1) (2) (3) (4) (5)

(B) (1) (2) (3) (4) (5)

(C) (1) (2) (3) (4) (5)

(D) (1) (2) (3) (4) (5)

For each entry in Column I, darken the bubbles of all the

matching entries. For example, if entry (A) in Column I, matches

with entries (2), (3) and (5), then darken these three bubbles in the

ORS. Similarly, for entries (B), (C) and (D).

Marking scheme:

For each entry in Column I

+2 If only the bubble(s) corresponding to all the correct match(es)

is(are) darkened



19. Match the Column:

Column I Column II

(A) In 2 ,R if the magnitude of

the projection vector of the

vector ˆ ˆα β+i j on ˆ ˆ3 +i j

is 3 and if α | 2 3β,= +

then possible value(s) of |α |

is (are)

1. 1

(B) Let a and b be real numbers

such that the function

2

2

3 2, 1( )

, 1

− − <=+ ≥ax x

f xbx a x

is

differentiable for all .∈x R

Then possible value(s) of a

is (are)

2. 2

(C) Let ω 1≠ be a complex

cube root of unity. If

2 4 3(3 3 2 ) +− +ω ω n

2 4 3(2 3 3 ) ++ + −ω ω n

2 4 3( 3 2 3 ) 0,++ − + + =ω ω n

then possible value(s) of n is

(are)

3. 3

(D) Let the harmonic mean of

two positive real numbers a

and b be 4. If q is a positive

real number such that a, 5,

q, b is an arithmetic

progression, then the

value(s) of | |−q a is (are)

4. 4

5. 5

Mathematics142

20. Match the thermodynamic processes given under Column I

with the expression given under Column II:

Column I Column II

(A) In a triangle let a, b

and c be the lengths of the

sides opposite to the angles

X, Y and Z, respectively. If

and

then

possible values of n for

which is (are)

1. 1

(B) In a triangle let a, b

and c be the lengths of the

sides opposite to the angles

X, Y and Z, respectively. If

then possible value(s) of

is (are)

2. 2

(C) In let

and be the

position vectors of X, Y and

Z with respect of the origin

O, respectively. If the

distance of Z from the

bisector of the acute angle

of with is

then possible value(s) of

is (are)

3. 3

(D) Suppose that denotes

the area of the region

bounded by x = 0, x = 2,

and

where Then the

value(s) of

when and is

(are)

4. 5

5. 6

,∆XYZ

2 2 22( )− =a b c

sin( ),

sin

−=λ

X Y

Z

cos( ) 0=πλn

,∆XYZ

1 cos2+ −X

2cos2 2sin sin ,=Y X Y

a

b

2 ,R ˆ ˆ ˆ ˆ3 , 3+ +i j i j

ˆ ˆ(1 )+ −β βi j

uuur

OXuuur

OY3

,2

| |β

(α)F

2 4=y x

| α 1 | | α 2 | α ,= − + − +y x x x

α 0,1.∈

8(α) 2,

3+F

α 0= α 1,=


143Mock Test-1


Time 3 Hours. Max. Marks 240 (80 for Mathematics)








4. Section 3 contains 2 “paragraph” type questions. Each paragraph describes an experiment, a situation or a problem. Two

multiple choice questions will be asked based on this paragraph. One or more than one option can be correct.



marking scheme too.






integer in the ORS.

Marking scheme:



1. Suppose that ,p qr r

and rr

are three non-coplanar vectors

in 3.R Let the components of a vector s

r

along ,p qr r

and

rr

be 4, 3 and 5, respectively. If the components of this

vector sr

along ( ),p q r− + +r r r

( )p q r− +r r r

and ( )p q r− − +r r r

are x, y and z, respectively, then the value of 2x y z+ + is

2. For any integer k, let cos sin ,7 7

k

k ki

π πα

= +

where

1.i = − The value of the expression

12

11

3

4 1 4 21

| |

| |

k kk

k kk

α α

α α

+=

− −=

Σ −

Σ −is

3. Suppose that all the terms of an arithmetic progression

(A.P.) are natural numbers. If the ratio of the sum of the

first seven terms to the sum of the first eleven terms is 6 :

11 and the seventh term lies in between 130 and 140, then

the common difference of this A.P. is

4. The coefficient of 9x in the expansion of 2(1 )(1 )x x+ +

3 100(1 )...(1 )x x+ + is

5. If the normals at the four points

and on the ellipse are concurrent, then

the value of must be

6. Let m and n be two positive integers greater than 1. If

cos( )

0lim

2

n

ma

e e eα

α→

− = −

then the value of m

nis

7. If

21

9 3tan

20

12 9( )

1

x x xe dx

xα

−+ +=

+ ∫ where

1tan x− takes

only principal values, then the value of 3

log |1 |4

e

πα

+

is

8. Let :f R R→ be a continuous odd function, which

vanishes exactly at one point and1

(1) .2

f = Suppose that

1( ) ( )

x

F x f t dt−

= ∫ for all [1 1,2]x∈ − and 1

( ) | ( ( )) |x

G x t f f t dt−

=∫

for all [ 1, 2].x∈ − If1

( ) 1lim ,

( ) 14x

F x

G x→= then the value of

1

2f

is

1 1 2 2 3 3( , ),( , ), ( , )x y x y x y

4 4( , )x y2 2

2 21

x y

a b+ =

1 2 3 4

1 2 3 4

1 1 1 1( )x x x x

x x x x

+ + + × + + +

Mathematics144







Marking scheme:


is(are) darkened.



9. Let

3

4

192( )

2 sin

xf x

xπ′ =

+for all x R∈ with

10.

2f

=

If

1

1/ 2( ) ,m f x dx M≤ ≤∫ then the possible values of m and M

are

a. 13, 24m M= = b. 1 1

,4 2

m M= =

c. 11, 0M M= − = d. 1, 12M M= =

10. Let S be the set of all non-zero real numbers α such that

the quadratic equation 2 0x xα α− + = has two distinct

real roots 1x and

2x satisfying the inequality 1 2| | 1.x x− <

Which of the following intervals is(are) a subset(s) of S ?

a. 1 1

,2 5

− −

b. 1

, 05

−

c. 1

0,5

d. 1 1

,25

11. If 1 6

3 sin11

α − =

and

1 43 cos ,

9β − =

where the inverse

trigonometric functions take only the principal values,

then the correct option(s) is (are)

a. cos 0β > b. sin 0β <

c. cos( ) 0α β+ > d. cos 0α <

12. Let 1E and

2E be two ellipse whose centers are at the

origin. The major axes of 1E and

2E lie along the x-axis

and the y-axis, respectively. Let S be the circle 2 2( 1) 2.x y+ − = The straight line 3x y+ = touches the

curves S, 1E and

2E at P, Q and R, respectively Suppose

that 2 2

.3

PQ PR= = If 1e and

2e are the eccentricities of

1E and 2 ,E respectively, then the correct expression(s)

is(are)

a. 2 2

1 2

43

40e e+ = b. 1 2

7

2 10e e =

c. 2 2

1 2

5| |

8e e− = d. 1 2

3

4e e =

13. Consider the hyperbola 2 2: 1H x y− = and a circle S with

center 2( , 0).N x Suppose that H and S touch each other at

a point 1 1( , )P x y with

1 1x > and 1 0.y > The common

tangent to H and S at P intersects the x-axis at point M. If

( , )l m is the centroix of the triangle ,PMN∆ then the

correct expression(s) is (are)

a. 2

1 1

11 for 1

3

dlx

dx x= − >

b.

( )1

12

11

for 13 1

xdmx

dx x= >

−

c. 12

1 1

11 for 1

3

dlx

dx x= + >

d. 1

1

1for 0

3

dmy

dy= >

14. The option(s) with the values of a and L that satisfy the

following equation is(are)

46 4

0

1 6 4

0

(sin cos )?

(sin cos )

at at dtL

e at at dt

π

π

+=

+

∫∫

a.

4 12,

1

ea L

e

π

π

−= =

− b.

4 12,

1

ea L

e

π

π

+= =

+

c.

4 14,

1

ea L

e

π

π

−= =

− d.

4 14,

1

ea L

e

π

π

+= =

+

15. Let , :[ 1, 2]f g R− → be continuous functions which are

twice differentiable on the interval (–1, 2). Let the values

of f and g at points –1, 0 and 2 be as given in the

following table:

1x = − 0x = 2x =

( )f x 3 6 0

( )g x 0 1 –1

In each of the intervals (–1, 0) and (0, 2) the function

( 3 )f g ′′− never vanishes. Then the correct statements(s) is

(are)

a. ( ) 3 ( ) 0f x g x′ ′− = has exactly three solutions in

( 1, 0) (0, 2)− ∪

b. ( ) 3 ( ) 0f x g x′ ′− = has exactly one solution in (–1, 0)

145Mock Test-1

c. ( ) 3 ( ) 0f x g x′ ′− = has exactly one solution in (0, 2)

d. ( ) 3 ( ) 0f x g x′ ′− = has exactly two solutions in (–1, 0)

and exactly two solutions in (0, 2)

16. Let 8 6 4 2( ) 7 tan 7 tan 3 tan 3 tanf x x x x x= + − − for all

,2 2

xπ π

∈ −

Then the correct expression(s) is (are)?

a. / 4

0

1( )

12xf x dx

π=∫ b.

/ 4

0( ) 0f x dx

π=∫

c. / 4

0

1( )

6xf x dx

π=∫ d.

/ 4

0( ) 1f x dx

π=∫


This section contains FOUR questions.


MORE THAN ONE of these four option(s) is(are) correct.



Marking scheme:


is(are) darkened.



Paragraph-I

Let :F R R→ be a thrice differentiable function. Suppose that

(1) 0, (3) 4F F= = − and ( ) 0F x′ < for all 1

, 32

x

∈

.

Let

( ) ( )f x xF x= for all .x R∈

17. The correct statement(s) is (are)

a. (1) 0f ′ <

b. (2) 0f <

c. ( ) 0f x′ ≠ for any (1, 3)x∈

d. ( ) 0f x′ = for some (1, 3)x∈

18. If 3

2

1( ) 12x F x dx′ = −∫ and

3

1( ) 40,F x dx′′ =∫ then the

correct expression(s) is(are)

a. 9 (3) (1) 32 0f f′ ′+ − =

b. 3

1( ) 12f x dx =∫

c. 9 (3) (1) 32 0f f′ ′− + =

d. 3

1( ) 12f x dx = −∫

Paragraph-II

Let 1n and

2n be the number of red and black balls,

respectively, in box I. Let 3n and

4n be the number of red and

black balls, respectively, in box II.

19. One of the two boxes, box I and box II, was selected at

random and a ball was drawn randomly out of this box. The

ball was found to be red. If the probability that this red ball

was drawn from box II is 1

,3

then the correct option(s) with

the possible values of 1 2 3, ,n n n and

4n is (are)

a. 1 2 3 43, 3, 5, 15n n n n= = = =

b. 1 2 3 43, 6, 10, 50n n n n= = = =

c. 1 2 3 48, 6, 5, 20n n n n= = = =

d. 1 2 3 46, 12, 5, 20n n n n= = = =

20. A ball is drawn at random from box I and transferred to

box II. If the probability of drawing a red ball from box I,

after this transfer, is 1

,3

then the correct option(s) with

the possible values of 1n and

2n is(are)

a. 1 24, 6n n= = b.

1 22, 3n n= =

c. 1 210, 20n n= = d.

1 23, 6n n= =


Mathematics146

ANSWER & SOLUTIONS

JEE-Main

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

d d c a a c c d a c

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

c d b a c c d a c c

21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

c c a d d a c c a d

1. (d) and

2. (d) If multiplying each term by the given

equation reduces to or or

which is not possible as considering Thus

given equation has no roots.

3. (c)

4. (b) It is understandable.

5. (a) Let be the roots of the equation

then Product of the roots

Since is a root of the equation.

6. (c) Trick: Put

Now by alternate (c),

put

7. (c)

8. (d) IUAENSRNC Obviously required number of words

are

9. (b) The probability that only two tests are needed =

probability that the first machine tested is faulty

probability that the second machine tested is faulty

10. (c)

.

11. (c) We know that

12. (d)

tanH d β= and tanH h d α− =

⇒ 60 tan

60 tanh

βα

=−

⇒ 60 tan 60 tan

tanh

α ββ

−− = ⇒

60sin( )

sincos cos

cos

hβ α

βα β

β

−=

⇒ cos sinx α β= .

13. (b) Since limit of a function is as therefore

to be continuous at a function, its value must be at

14. (a) 4siny a θ=

⇒ 34 sin cos

dya

dθ θ

θ=

and 4cosx a θ=

⇒ 34 cos sin

dxa

dθ θ

θ= −

∴ 2

2

2

/ sintan

/ cos

dy dy d

dx dx d

θ θθ

θ θ−

= = = −

∴ 2

3

4

3tan 1.

4

dy

dx πθ

π

=

= − = −

3 0x + > 2 3 2 0x x+ + ≠

1,x ≠ ( 1),x−

( 1) ( 1)x x x− = − 2( 1) 0x − =

1,x = 1.x ≠

2 2 2 2(3 5 3 ) (3 3 5 )ω ω ω ω+ + + + +2 2 2 2 2(3 3 3 2 ) (3 3 3 2 )ω ω ω ω ω ω= + + + + + + +

2 3(1 0, 1)ω ω ω+ + = =2 2 2 2 4(2 ) (2 ) 4 4 4( 1) 4.ω ω ω ω= + = + = − = −

, ,A

A ARR

3 2 0ax bx cx d+ + + =

3A =

d

a= −

⇒1/3

dA

a

= −

A

∴ 3 2 0aA bA cA d+ + + =

⇒2 /3 1/3

0d d d

a b c da a a

− + − + − + =

⇒2 /3 1/3

d db c

a a

=

⇒2

3 3

2

d db c

a a= ⇒ 3 3 .b d c a=

1, 2, 3,....n =

1 21, 2 2 4S S= = + =

1, 2n =0 1

1 21.2 1, 2.2 4S S= = = =

2 4 log log(log ) (log )1

2! 4! 2

e en n

e en n e e

−++ + + =K

1

.2

n n−+=

6 !4 ! 8640

2 !× =

×

2 1 1

4 3 6= × =

3sin sin sin18 .sin 54

10 10

π π= ° °

5 1 5 1 1sin18 .cos36 .

4 4 4

− += ° ° = =

( )1 2cosh log 1x x x− = + −

∴ ( )1 2cosh (1) log 1 1 1 log1 0.− = + − = =

a b+ 0,x →

a b+

0x =

(0) .f a b⇒ = +

H=60m α

β

d

h

147Mock Test-1

15. (c) Given

⇒

⇒

Now

Hence minimum value at

16. (c) Putting we get

17. (d) Area

18. (a)

Now integrating both sides, we get

19. (c) 4 3 11x y+ = and 15

4 32

x y+ =

Therefore, 15

1172

5 10D

−= =

.

20. (c) Since area where

Area

21. (c) Let be the mid-point of the line segment

joining the focus and a general point on the

parabola.

Then

Put these values of x and y in we get

So, locus of is

Its directrix is

22. (c) Hence it is parallel to

23. (a) Direction ratio of the line joining the point

are

⇒

Direction ratio of the line parallel to line

are

Angle between two lines,

⇒

24. (d) Since

25. (d)

Hence, number of solutions is zero.

[(5 )(2 )]( )

[1 ]

x xf x

x

+ +=

+

4 4( ) 1 (5 ) (6 )

1 (1 )f x x x

x x= + + + = + +

+ +

2

4'( ) 1 0;

(1 )f x

x= − =

+2 2 3 0x x+ − =

3,1x = −

3

8( ) ,

(1 )f x

x′′ =

+

( 3) ,f ve′′ − = −

(1)f ve′′ = +

1x =

(5 1)(2 1) 6 3(1) 9.

(1 1) 2f

+ + ×= = =

+

1

2

1tan ,

1t x dt dx

x

−= ⇒ =+

1tan

21

xte

dx e dtx

−

=+∫ ∫

1tan .t xe c e c

−

= + = +

0log .A x dx

∞= ∫

0( log )x x x ∞= − = ∞

2 2( )x y y y xdye x e e e x

dx

− − −= + = +

⇒ 2( )y xe dy x e dx= +

3

.3

y xxe e c= + +

2,rπ=

r a=

⇒ 2 .aπ=

( , )P h k

( ,0)a ( , )Q x y

, 2 , 2 .2 2

x a yh k x h a y k

+= = ⇒ = − =

2 4 ,y ax=

24 4 (2 )k a h a= −

⇒ 2 2 2 24 8 4 2k ah a k ah a= − ⇒ = −

( , )P h k 2 22y ax a= −

⇒ 2 22

ay a x

= −

0.2 2

a ax x− = − ⇒ =

ˆ ˆ ˆ ˆ3 9 3( 3 ).a b i j i j+ = + = +r

r

).3,1(

(2, 1, 3), ( 3, 1, 7)− − 1 1 1( , , )a b c

( 3 2, 1 1, 7 ( 3))− − − − −

⇒ ( 5, 0, 10)−

1 3

3 4 5

x y z− += =

2 2 2( , , )a b c

⇒ (3, 4, 5)

1 2 1 2 1 2

2 2 2 2 2 2

1 1 1 2 2 2

cosa a b b c c

a b c a b cθ

+ +=

+ + + +

( 5 3) (0 4) (10 5)cos

25 0 100 9 16 25θ

− × + × + ×=

+ + + +

35cos

25 10θ =

⇒ 1 7cos

5 10θ − =

qpqp ~)(~ ∧≡⇒

qpqp ~~)(~~ ∧=⇒

2 1 1

1 3 2

1 4 3

−

∆ = −

−

2(9 8) 1( 3 2) 1(4 3) 7 7 0= − − − − − + = − =

y =loge x

(1,0) X

Y

Mathematics148

26. (a)

27. (c) For continuity at all we must have

. . .(i)

and

. . .(ii)

From (i) and (ii),

and .

28. (c) Put then

29. (a) Equation of line perpendicular to

is

Now it is a tangent to the circle, if

Radius of circle

Distance of line from centre of circle

or

Hence equations of tangents are

and

30. (d) .

JEE Advance Paper-I

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

4 8 4 8 5 1 0 4 a,c,d c, d

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

b,c b,d a,b a,d a,c b,c a,d a,b,c c c

1. (4)

Required area sq unit

2. (8)2 4 4 6 65

cos 2 cos sin cos sin 24

x x x x x+ + + + =

⇒ 2 2 25

cos 2 5cos sin 04

x x x− =

⇒ 2tan 2 1, where 2 [0, 4 ]x x π= ∈

Number of solutions = 8

3. (4) Image of 5y = − about the line 4 0x y+ + = is

1x =

⇒ Distance 4AB =

4. (8) Let coin was tossed ‘n’ times

Probability of getting at least two heads 1

12 2n n

n = − +

⇒ 1

1 0.962n

n + − ≥

⇒ 2

251

n

n≥

+⇒ 8n ≥

5. (5) .6! 5!n = (5 girls together arranged along with 5

boys)4 .(7! 2.6!).4!m C5= −

(4 out of 5 girls together arranged with others – number of

cases all 5 girls are together)

5 5 6! 4!

6! 5!

m

n

⋅ ⋅ ⋅=

⋅

2 3 4

1 1 1 1

2 3 4n n n n− + − +K

2 3 41 (1/ ) (1/ ) (1/ )

2 3 4

n n n

n= − + − +K

1 1log 1 log .e e

n

n n

+ = + =

,x R∈

( / 2)lim ( 2sin )

2 xf x

π

π−→ −

− = −

( / 2)lim ( sin )

xA x B

π +→ −= +

⇒ 2 A B= − +

( / 2)lim ( sin )

2 xf A x B

π

π−→

= +

( / 2)lim (cos )

xx

π +→=

⇒ 0 A B= +

1−=A

1=B

4 33 5 20 ,x t x dx dt+ = ⇒ =

3 4 1/ 213 5

20x x dx t dt+ =∫ ∫

3/ 2 4 3/ 22 1 1. (3 5 ) .

3 20 30t c x c= × + = + +

5 12 8 0x y+ + =

12 5 0.x y k− + =

=

12(11) 5(2)121 4 25

144 25

k− ++ − =

+

⇒ 8k = 252.−

12 5 8 0x y− + =

12 5 252x y− =

)(~~ pqqp →=∧

4 (1 1) 4= × × =

∴ 4λ =1 1 1 1 1/ 2

2 4 8 16 1 1/ 2 4λ λ λ λ λ λ λ+ + + + ∞

−∞ = = = =K

K

O

7

6

5

4

3

2

1 –1 –2

–1

–2

–3

1 2 3 4 5

149Mock Test-1

6. (1) Let Parabola be and coordinates of P and Q

on this parabola are and T

is the point of intersection of tangents at and

Coordinates of

Similarly ,

Let

∴

or

Similarly,

or

7. (0)

0 1 2

1 0 1

0 0 1 10

2 0 2 1 2 0 4

x x xI dx dx dx

−

⋅ ⋅ ⋅= + + + =

+ + +∫ ∫ ∫

⇒ 4 1 0I = =

8. (4) Let inner radius be r and inner length be l

2r Vπ =l

Volume of material be M,

2 2( 2) ( 2)M r rπ π= + + −l l

2 3

4 88 0 4

dM V Vr

dr r rπ π= − − + + +

0 when 10dM

rdr

= =

⇒ 1000V π=

⇒ 4250

V

π=

9. (a,c,d) | | | |b c a+ =r

r r

⇒ 2 2 2| | | | 2 | |b c b c a+ = ⋅ =

r r

r r r

⇒ 248 | | 48 144c+ + =r

⇒ | | 4 3c =r

∴2| |

| | 122

ca− =

r

r

Also, | | | |a b c+ =r

r r

⇒ 2 2 2| | | | 2 | |a b a b c+ = ⋅ =

r r

r r r

⇒ 72a b⋅ = −r

r

0a b c+ + =r

r r

⇒ a b c a× = ×r

r r r

⇒ | | 2 | | 48 3a b c a a b× + × = × =r r

r r r r

10. (c,d) 3 4 4 3( )TY Z Z Y− 4 3 3 4( ) ( ) ( ) ( )T T T TZ Y Y Z= −

4 3 3 4Z Y Y Z= − +

⇒ symmetric 44 44 is symmetricX Y+

4 3 3 4 skew symmetricX Z Z X−

23 23 skew symmetric.X Y+

11. (b,c) We get

2 2 2(1 α) (1 2α) (1 3α)

3 2α 3 4α 3 6α

5 2α 5 4α 5 6α

+ + +

+ + +

+ + +

3 3 2 2 2 1648α ( ; )R R R R R R= − → − → −

2 2 2α 2 4α 2 9α 2

3 2α 3 4α 3 6α

2 2 2

− − −

+ + +

1 1 2 3 3 2648α ( ; )R R R R R R= − → − → −

⇒

2 2 22α 5α 9α 3

2α 2α 3 6α 648α

0 0 2

− − − −

− − + = −

12. (b,d) Let the required plane be 1 0x z yλ+ + − =

⇒ 2

| 1 | 11

22

λλ

λ

−= ⇒ = −

+

⇒ 3 2 2 2 0P x y z≡ − + − =

Distance of 3P from ( , , )α β γ is 2

| 2 2 2 |

24 1 4

α β γ− + −=

× +

⇒ 2 2 4 0 and 2 2 8 0α β λ α β λ− + + = − + − =

13. (a,b) Line L will be parallel to the line of intersection of

1P and 2P

Let a, b and c be the direction ratios of line L

⇒ 2 0 and 2 0a b c a b c+ − = − + =

⇒ : : ::1 : 3 : 5a b c − −

Equation of line L is 0 0 0

1 3 5

x y z− − −= =

− −

Again foot of perpendicular from origin to plane 1P is

1 1 1, ,

6 3 6

− −

∴ Equation of project of line L on plane 1P is

1 2 1

6 6 6

1 3 5

x y z

k

+ + −= = =

− −

Clearly points 5 2

0, ,6 3

− −

and 1 1 1

, ,6 3 6

− −

satisfy the

line of projection i.e. M

2 4y ax=2

1 1( , 2 )P at at≡ 2

2 2( , 2 );Q at at≡

1t 2.t

∴ 1 2 1 2 , , ( )T at t a t t≡ +

3 1 3 1 , , ( )P at t a t t′ ≡ +

2 3 2 3 , , ( )Q at t a t t′ ≡ +

: : 1TP TP λ′ =

3 2

1 2

t t

t tλ

−=

−

3 2

1 2

t tTP

TP t t

′ −=

−

1 3

1 2

t tTQ

TQ t t

′ −=

−

1TP TQ

TP TQ

′ ′= =

Mathematics150

14. (a,d) 2( , 2 )P at at ⇒ 2

16 8,

a aQ

tt

−

1

2OPQ OP OQ∆ = ⋅

⇒ 2

2

1 (4) 164 4 3 2

2

aat t

t t+ ⋅ + =

2 3 2 4 0t t− + =

⇒ 2, 2 2t =

Hence, 2

2( , 2 ) ,2

tP at at P t

=

2t = ⇒ (1, 2)P

2 2t = ⇒ (4,2 2)P

15. (a,c)1

1 1

x

x x

dy ye

dx e e+ =

+ +

In(1 )

. . 11

xx

e x

x

e

eI F dx e e

e

+= = = ++∫

⇒ (1 ) 1xy e dx+ = ∫ ⇒1 x

x cy

e

+=

+

(0) 2y =

⇒ 1c =

⇒ 4

1 x

xy

e

+=

+

⇒ ( 4) 0y − =

⇒ 2

(1 ) ( 4)0

(1 )

x x

x

e x ey

e

+ − +′ = =

+

2

(1 ) ( 4)Let ( )

(1 )

x x

x

e x eg x

e

+ − +=

+

2

2 4(0) 0

2g

−= <

2 2

1 3 21 1

( 1) 0 01 1

1 1

e e eg

e e

+ − − − = < = > + +

(0) ( 1) 0.g g⋅ − < Hence g(x) has a root in between ( 1, 0)−

16. (b,c) Let the family of circles be 2 2 α α 0x y x y c+ − − + =

On differentiation 2 2 α α 0x yy y′ ′+ − − =

Again on differentiation and substituting 'α' we get

2 2 2

2 2 2 01

x yyx y yy y

y

′ +′ ′′ ′′+ + − = ′+

⇒ 2( ) (1 ) 1 0y x y y y y′′ ′ ′ ′− + + + + =

17. (a,d) Differentiability of ( )f x at 0x =

0 0

(0) (0 ) 0 ( )LHD (0 ) lim lim 0

f f gf

δ δ

δ δδ δ

−

→ →

− − + − ′ = = =

0 0

(0 ) (0) ( )RHD (0 ) lim lim 0

f f gf

δ δ

δ δδ δ

+

→ →

+ − ′ = = =

⇒ ( )f x is differentiable at 0x =

Differentiabiligy of ( )h x at 0x =

(0 ) 1, ( )h h x+′ = is an even function

Hence non diff. at 0x =

Differentiability of ( ( ))f h x at 0x =

| |( ( )) ( )xf h x g e x R= ∀ ∈

0

( (0)) ( (0 ))LHD ( (0 )) lim

f h f hf h

δ

δδ

−

→

− −′ =

0

(1) ( )lim (1)

g g eg

δ

δ δ→

−′= =

0

( (0 )) ( (0))RHD ( (0 )) lim

f h f hf h

δ

δδ

+

→

+ −′ =

0

( ) (1)lim (1)

g e gg

δ

δ δ→

−′= = Since (1) 0g ′ ≠

⇒ ( ( ))f h x is non diff. at 0x =

Differentiability of ( ( ))h f x at 0x =

( ( )|, 0( ( ))

1, 0

f xe xh f x

x≠==

0

( (0)) ( (0 ))LHD. ( (0 )) lim

h f h fh f

δ

δδ

δ→

− −′ − =

| ( )|

0

1 | ( ) |lim 0

| ( ) |

ge g

g

δ

δ

δδ δ

−

→

− −= ⋅ =

−

18. (a,b,c) Given ( ) sin2

g x x x Rπ

= ∀ ∈

1

( ) sin ( ( ))3

f x g g x =

Also, ( ( ( ))) ,2 2

g g g x x Rπ π ∈ − ∀ ∈

Hence, 1 1

( ) and ( ( )) ,2 2

f x f g x ∈ −

0 0

1 1sin ( ( )) ( ( ))( ) 3 3lim lim

1( ) ( )( ( ))

3

x x

g g x g g xf x

g x g xg g x

→ →

= ⋅

⇒ 0

sin sin2

lim6 6

sin2

x

x

x

ππ π

π→

⋅ =

151Mock Test-1

Range of 1 1

( ( )) sin , sin2 2 2 2

g f xπ π ∈ −

⇒ ( ( )) 1.g f x ≠

19. (c) (A) 3

32

α β+=

3 2 3α β+ =± . . .(i)

Given 2 3α β= + . . .(ii)

From equation (i) and (ii), we get α 2 1or= −

So |α| 1 or 2=

(B) 2

2

13 2,( )

1,

xaxf x

xbx a

<− −=

≥+

For continuity 23 2a b a− − = +

2 3 2a a b+ + = − . . .(i)

For differentiability 6a b− =

6a b= −

2 3 2 0a a− + =

1, 2a =

(C) 2 4 3 2 4 3(3 3 2 ) (2 3 3 )n nω ω ω ω+ +− + + + −

2 4 3( 3 2 3 ) 0nω ω ++ − + + =

2 4 3 2 4 3(3 3 2 ) ( (2 3 3 ))n nω ω ω ω ω+ +− + + + −

2 2 4 3( ( 3 2 3) 0nω ω ω ++ − + + =

⇒ 2 4 3 4 8(3 3 2 ) (1 ) 0n n nω ω ω ω+− + + + + =

⇒ 3 ,n k k N≠ ∈

(D) Let 5a d= −

5q d= +

5 2b d= +

⇒ | | | 2 |q a d− =

Given 2

4ab

a b=

+

⇒ 2ab

a b=

+

(5 )(5 2 ) 2(5 5 2 ) 2(10 )d d d d d− + = − + + = +

225 10 5 2 20 2d d d d+ − − = +

22 3 5 0d d− − =

5

1,2

d d= − =

⇒ | 2 | 2,5d =

20. (c) (A) 2

2 2 (given)2

ca b− =

2

2 2 2 244 (sin sin ) sin ( )

2

RR X Y Z− =

⇒ 22(sin( ) sin( ) sin ( )X Y X Y Z− ⋅ + =

⇒ 22 sin( ) sin( ) sin ( )X Y Z Z⋅ − ⋅ =

⇒ sin( ) 1

sin 2

X Y

Zλ

−= =

⇒ cos 0 for n odd integer.2

nπ = =

(B) 1 cos 2 2 cos 2 2 sin sinX Y X Y+ − =

2 2sin sin sin 2sin 0X X Y Y+ − =

(sin sin )(sin 2sin ) 0X Y X Y− + =

⇒ sin sinX Y=

⇒ sin

1.sin

X a

Y b= =

(C) Here, distance of Z from bisector of

3

and2

OX OY =uuur uuur

⇒

2 21 1 9

2 2 2β β − + − =

⇒ 2, 1β = −

⇒ | | 2,1β =

(D) When α 0=

2

0Area 6 2 x dx= − ∫

8 26

3= −

When α 1=

1 2

0 1Area (3 2 ) ( 1 2 )x x dx x x dx= − − + + −∫ ∫

1 22 2

3/ 2 3/ 2

0 1

4 43

2 3 2 3

x xx x x x= − − + + −

8

5 2.3

= −


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

9 4 9 8 4 2 9 7 d a,d

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

b,c,d a,b a,b,c a,c b,c a,b a,b,c c,d a,b c,d

1. (9) 4 3 5= + +r r r r

s p q r

( ) ( ) ( )= − + + + − + + − − +r r r r r r r r r r

s x p q r y p q r z p q r

( ) ( ) ( )= − + − + − − + + +r r r r

s x y z p x y z q x y z r

⇒ 4− + − =x y z

Mathematics152

⇒ 3− − =x y z

⇒ 5+ + =x y z

On solving we get 9 7

4, ,2 2

= = = −x y z

⇒ 2 9+ + =x y z

2. (4)

127 7

1

3(4 2) 7

1

112

43

1

π π

π

=

−

=

Σ −

= =

Σ −

ki i

k

ii k

k

e e

e e

3. (9) Let seventh term be ‘a’ and common difference be ‘d’

Given 7

11

6

11=

S

S ⇒ 15=a d

Hence, 130 15 140< <d ⇒ 9=d

4. (8)9x can be formed in 8 ways

i.e., 9 1 8 2 7 3 6 4 5 1 2 6 1 3 5 2 3 4, , , , , , ,+ + + + + + + + + +x x x x x x x x and

coefficient in each case is 1

⇒ Coefficient of 9

8 times1 1 1 ......... 1 8= + + + + =x

5. (4) Let point of concurrent is

Equation of normal at is,

It is passes through , then

. . .(i)

But or . . .(ii)

Value of from equation (ii), putting in equation (i),

we get

Arranging above as a fourth degree equation in we get

Above equation being of fourth degree in therefore

roots of the above equation are then

. . .(iii)

. . .(iv)

Multiplying equation (iii) and (iv), we get

6. (2)

( )cos

0lim

2

α

α α→

−= −

n

m

e e e

( )( )

(cos( ) 1)

20

(cos 1)lim

cos( ) 1

α

α

α

α α α

−

→

−

−

nn

n m n

e e 2

2α = −n e

if and only if 2 0− =n m

7. (9)1

21

(9 3tan )

20

12 9

1α

−+ +=

+ ∫ x x x

e dxx

Put 19 3tan−+ =x x t

⇒ 2

39

1

+ = +

dx dtx

⇒ 33

9944

01

ππ

α++

= = −∫ te dt e

⇒ 3

log |1 | 94

πα

+ − =

e

8. (7) 1

1(1) | ( ( )) | 0

−= =∫G t f f t dt ( ) ( )− = −f x f x

Given 1

(1)2

=f

1 1

( ) (1)

( ) (1) 11lim lim( ) (1)( ) | ( (1)) | 14

1

→ →

−−= = =−−

x x

F x F

F x fxG x GG x f f

x

⇒ 1/ 2 1

| (1/ 2) | 14=

f

⇒1

7.2

=

f

9. (D) 3 3

1/ 2 1/ 2

192 192( )

3 2≤ ≤∫ ∫

x x

t dt f x t dt

4 4 316 1 ( ) 24

2− ≤ ≤ −x f x x

1 1 1

4 2

1/ 2 1/ 2 1/ 2

3(16 1) ( ) 24

2

− ≤ ≤ −

∫ ∫ ∫x dx f x dx x dx

1

1/ 2

26 391 ( ) 12

10 10< ≤ ≤ <∫ f x dx

10. (a,d) Here, 2

1 20 ( ) 1< − <x x

⇒ 2

1 2 1 20 ( ) 4 1< + − <x x x x

⇒ 2

10 4 1

α< − <

⇒ 1 1 1 1

, ,2 25 5

α

∈ − − ∪

( , )h k

( , )x y′ ′ 1

2 2/ /

y yx x

x a y b

′′ −−=

′ ′

( , )h k

2 2 2 2 4 2 2 ( ) y a h x b x b k x′ ′ ′ ′− + =

2 2

2 21

x y

a b

′ ′= =

22 2 2

2( )

by a x

a′ ′= −

2y′

22 2 2 2 2 2 4 2 2

2( ) ( )

ba x a h b a x b k x

a′ ′ ′− + − =

⇒2

2 2 4 2 2 2 2 2 2 2 2 4 2 2

2( ) ( ) 2 ( )

ba x a h b a x a hx b a b k x

a′ ′ ′ ′− + − + − =

,x′

⇒ 2 2 2 4 2 2 2 3 2 4 2 2 6 2( ) 2 ( ) ( ) 2 ( ) 0a b x ha a b x x a h a b x a h′ ′ ′ ′− − + − + − − + =K

,x′

1 2 3 4, , ,x x x x

2 2 2 2

1 2 3 4 2 2 2 2 2

2 ( ) 2( )

( ) ( )

ha a b hax x x x

a b a b

−+ + + = − =

− − −

1 2 3

1 2 3 4 1 2 3 4

1 1 1 1 x x x

x x x x x x x x

Σ+ + + =

⋅ ⋅ ⋅ 4 2 2

2 22 2 2

6 2 2

2 2 2

2 ( )

2( )( )

( )

a h a b

a ba b

a h a h

a b

−−− −

= =

− −

1 2 3 4

1 2 3 4

1 1 1 1( ) 4x x x x

x x x x

+ + + × + + + =

153Mock Test-1

11. (b,c,d)3

,2 2

π πα π π β< < < <

⇒ 3 5

2 2

π πα β< + <

⇒ sin 0; cos 0β α< <

⇒ cos( ) 0.α β+ >

12. (a,b) For the given line, point of contact for

2 2

1 2 2: 1+ =

x yE

a b is

2 2

,3 3

a b

and for

2 2

2 2 2: 1+ =

x yE

B A is

2 2

,3 3

B A

Point of contact 3+ =x y of and circle is (1, 2)

Also, general point on 3+ =x y can be taken as

1 , 22 2

±

m

r r where,

2 2

3=r

So, required points are 1 8

,3 3

and 5 4

,3 3

Comparing with points of contact of ellipse,

2 25, 8= =a B

2 24, 1= =b A

∴ 1 2

7

2 10=e e and 2 2

1 2

43

40+ =e e

13. (a,b,d) Tanget at P, 1 1 1− =xx yy intersects x-axis at

1

1, 0

Mx

Slope of normal 1 1

1 1 2

0−= − =

−y y

x x x

⇒ 2 1 12 (2 ,0)= ⇒ ≡x x N x

For centroid 1

1 1

13

,3 3

+= =l

xx y

m

2

1 1

11

3= −

ld

dx x

1 1

21 1 1 1

1 1,

3 3 3 1= = =

−

dy xdm dm

dy dx dx x

14. (a,c) Let 1 6 4

0(sin at cos at)

π+ =∫ e dt A

2

1 6 4(sin at cos at)π

π= + +∫I e dt

Put π= +t x

=dt dx

for 2=a as well as 4=a

Similarly

3

1 6 4 2

2(sin at cos at)

π π

π+ =∫ e dt e A

So,

2 3 43 1

1

π π π π

π

+ + + −= =

−A e A e A e A

LA e

For both 2, 4=a

15. (b,c) Let ( ) ( ) 3 ( )= −H x f x g x

( 1) (0) (2) 3.− = = =H H H

Applying Rolle’s Theorem in the interval [–1, 0]

( ) ( ) 3 ( ) 0′ ′ ′= − =H x f x g x for atleast one ( 1, 0)∈ −c

As ( )′′H x never vanishes in the interval

⇒ Exactly one ( 1,0)∈ −c for which ( ) 0′ =H x

Similarly, apply Rolle’s Theorem in the interval [0, 2]

⇒ ( ) 0′ =H x has exactly one solution in (0, 2)

16. (a,b) 6 2 2( ) (7 tan 3 tan ) (tan 1)= − +f x x x x

/ 4 / 4

6 2 2

0 0( ) (7 tan 3tan )sec

π π= −∫ ∫f x dx x x xdx

⇒ / 4

0( ) 0

π=∫ f x dx

/ 4/ 4 / 4

0 00( ) ( ) ( )

ππ π = − ∫ ∫ ∫ ∫xf x dx x f x dx f x dx dx

/ 4

0

1( )

12

π=∫ xf x dx

17. (a,b,c) (a) ( ) ( ) ( )′ ′= +f x f x xF x

(1) (1) (1)′ ′= +f F F

(1) (1) 0′ ′= <f F

(1) 0′ <f

(b) (2) 2 (2)=F F

( )F x is decreasing and (1) 0=F

Hence (2) 0<F

⇒ (2) 0<f

(c) ( ) ( ) ( )′ ′= +f x F x xF x

( ) 0 (1, 3)< ∀ ∈F x x

( ) 0 (1, 3)′ < ∀ ∈F x x

Hence ( ) 0 (1, 3)′ < ∀ ∈f x x

Mathematics154

18. (c,d)3 3

1 1( ) ( )=∫ ∫f x dx xF x dx

32

32

11

1( ) ( )

2 2

′= −

∫

xF x x F x dx

9 1(3) (1) 6 12

2 2= − + = −F F

3

3 3 2

11

40 [ ( )] 3 ( )′ ′= − ∫x F x x F x dx

40 27 (3) (1) 36′ ′= − +F F . . .(i)

( ) ( ) ( )′ ′= +f x F x xF x (3) (3) 3 (3)′ ′= +f F F

(1) (1) (1)′ ′= +f F F 9 (3) (1) 32 0′ ′− + =f f

19. (a,b) P (Red Ball) (1) ( | ) ( ) ( | )= ⋅ + ⋅P P R I P II P R II

1 ( ) ( | )

( | )3 ( ). ( | ) ( ) ( | )

⋅= =

+ ⋅P II P R II

P II RP I P R I P II P R II

3

3 4

31

1 2 3 4

1

3

+=

++ +

n

n n

nn

n n n n

of the given options (a) and (b) satisfy above condition

20. (c,d) P (Red after Transfer) = P (Red Transfer), P (Red

Transfer in II Case)

+ P (Black Transfer). P (Red Transfer in II Case)

1 1

1 2 1 2

( 1)( )

( 1)

−= +

+ + −

n nP R

n n n n

2 2

1 2 1 2

1.

1 3=

+ + −

n n

n n n n

of the given option 3 and 4 satisfy above condition

155Mock Test-2


1. If is given by then

equals:

a. b.

c. d.

2. Let and be the roots of the equation

The equation whose roots are is

a. b.

c. d.

3. Let then

is equal to

a. 0

b.

c.

d.

4. The value of is

a. –1 b. 0

c. –i d. i

5. The inverse of the matrix is

a. b.

c. d.

6. The sum of the first five terms of the series

will be

a. b.

c. d.

7. If then the

expression has the value

a. 32 b. 63


8.

a. b. c. d.

9. is

equal to

a. b.

c. d.

10. There are 10 lamps in a hall. Each one of them can be

switched on independently. The number of ways in which

the hall can be illuminated is.

a. b. 1023

c. d. 10 !

11. A fair coin is tossed repeatedly. If tail appears on first four

tosses, then the probability of head appearing on fifth toss

equals

a. b.

c. d.

12. If then

a. 2 b. 4 c. 8 d. 16

13. If and

,then is equal to

a. b. c. d.

14. Find imaginary part of

a. b.


15. An observer on the top of a tree, finds the angle of

depression of a car moving towards the tree to be 30°.

After 3 minutes this angle becomes 60°. After how much

more time, the car will reach the tree

a. 4 min b. 4.5 min c. 1.5 min d. 2 min

: [1, ) [2, )f ∞ → ∞1

( )f x xx

= +

1( )f x−

2 4

2

x x+ −21

x

x+

2 4

2

x x− − 21 4x+ −

α β 2 1 0x x+ + =19 7,α β

2 1 0x x− − = 2 1 0x x− + =2 1 0x x+ − = 2 1 0x x+ + =

22 2cos sin , 1,n i i

n n

π πω = + = −

2

3 3( )x y zω ω+ + 2

3 3( )x y zω ω+ +

2 2 2x y z+ +2 2 2x y z yz zx xy+ + − − −2 2 2x y z yz zx xy+ + + + +

6

1

2 2sin cos

7 7k

k ki

π π

=

−

∑

1 0 0

0 1 0

0 0 1

0 0 1

0 1 0

1 0 0

1 0 0

0 1 0

0 0 1

0 1 0

0 0 1

1 0 0

1 0 0

0 0 1

0 1 0

1 33 4 6 ......

2 4+ + +

939

16

318

16

739

16

913

16

2 6 2 12

1 2 12(1 2 ) 1 .... ,x x a x a x a x+ − = + + + +

2 4 6a a a+ + ....+

12a+

1 2 1 2 3 1 2 3 41

2! 3! 4!

+ + + + + ++ + + + ∞ =K

e 3e / 2e 3 / 2e

1 1 1log 2 log 1 log 1 .... log 1

2 3 1e e e e

n

+ + + + + + + −

log 1e

loge

n

log (1 )e

n+ log (1 )e

n−

210

102

1

2

1

32

31

32

1

5

22

2

tan 60 cosec30sin 45 cos 60 ,

sec 45 cot 30x

° °° ° =

° °x =

sin sin 2 sin3 sinθ θ θ α+ + =

cos cos 2 cos3 cosθ θ θ α+ + = θ

/ 2α α 2α / 6α

1 5 7 9sin

16

i− −

log 2 log 2−

Mathematics156

16. Let If be

continuous for all x, then

a. 7 b. –7

c. d. None of these

17. If ....

,x

x ex ey e

+ ∞++= then dy

dx=

a. 1

y

y− b.

1

1 y−

c.

1

y

y+ d.

1

y

y −

18. One maximum point of sin cosp qx x is

a. 1tan ( / )x p q

−= b. 1tan ( / )x q p

−=

c. 1tan ( / )x p q−= d. 1tan ( / )x q p−=

19.

a. b.

c. d.

20. Area bounded by the parabola and the ordinates

is

a. b.

c. d. None of these

21. The solution of the differential equation is

a. b.

c. d.

22. The solution of the differential equation

is

a. b.

c. d. None of these

23. The vertex of an equilateral triangle is (2,–1) and the

equation of its base in The length of its sides

is

a. b.

c. d.

24. If the equation 2 2( 1) ( 2)

13 4

K x y+ ++ = represents a circle,

then K =

a. 3/4 b. 1 c. 4/3 d. 12

25. The equation of the common tangent to the curves

and is

a. b.

c. d.

26. The vectors and are non-collinear. The value of x for

which the vectors and

are collinear, is

a. 1 b.

c. d. None of these

27. If vectors satisfy the condition

then is equal to

a. 0 b. –1 c. 1 d. 2

28. The angle between the straight lines

and is

a. b. c. d.

29. A variable plane is at a constant distance p from the origin

and meets the axes in A, B and C. The locus of the

centroid of the tetrahedron is

a. b.

c. d. None of these

30. is equal to

a. b.

c. d.

3 2

2

16 20,if 2

( ) .( 2), if 2

x x xx

f x xk x

+ − + ≠= − =

( )f x

k =

7±

2

1

(log )dx

x x=∫

1

logc

x+

1

logc

x− +

log log x c+ log log x c− +

22y x=

1, 4x x= =

4 2.

3sq unit

28 2.

3sq unit

56

3sq. unit

210

dy x

dx x

++ =

11tan

2y x c−= − +

2

log 02

xy x c+ + + =

11tan

2y x c−= +

2

log2

xy x c− − =

cos log(sec tan ) cos logy x x dx x+ = (sec tan )y y dy+2 2sec secx y c+ = sec secx y c+ =

sec secx y c− =

2 1.+ =x y

4 / 15 2 / 15

4 / 3 3 1/ 5

2 8y x= 1xy = −

3 9 2y x= + 2 1y x= +

2 8y x= + 2y x= +

ar

br

( 2)c x a b= − +rr r

(2 1)d x a b= + −r rr

2

1

3

1

,, ca brrr

| | | |,a c b c− = −rr r r

( )2

a bb a c

+− ⋅ −

rrr r r

1 2 3

2 5 4

x y z+ − += =

1

1

x −=

2

2

y + 3

3

z −=

−°45 °30 °60 °90

OABC2 2 2 2

16x y z p− − − −+ + = 2 2 2 1

16x y z p− − − −+ + =

2 2 216x y z

− − −+ + =

~ ( )∨p q

~ ~∨p q ~ ~∧p q

~ ∨p q ~∨p q


157Mock Test-2

JEE ADVANCE PAPER-I



9 (both inclusive).

1. The coefficient of in the expansion of

is

2. The minimum value of the sum of real numbers

5 4 3 8a , a , 3a , 1, a− − − and 10a with a 0> is

3. The value of

3

2

1 1 1 16 log 4 4 4 is

3 2 3 2 3 2 3 2

+ − − −

4. The number of distinct solutions of the equation

2 4 45cos 2 cos sin

4x x+ + 6cosx + 6sin 2x x+ = in the

interval is

5. A cylindrical container is to be made from certain solid

material with the following constraints: It has a fixed

inner volume of 3mm ,V has a 2 mm thick solid wall and

is open at the top. The bottom of the container is a solid

circular disc of thickness 2 mm and is of radius equal to

the outer radius of the container.

If the volume of the material used to make the container is

minimum when the inner radius of the container is 10

mm, then the value of is

6. Let the curve C be the mirror image of the parabola

with respect to the line If A

and B are the points of intersection of C with the line

then the distance between A and B is

7. Let be a continuous odd function, which vanishes

exactly at one point and Suppose that

for all and

for all If then the value of is

8. Consider the set of eight vectors

Three non-coplanar vectors

can be chosen from ways. Then p is.



9. Let f be a non-negative function defined on the interval

1[0, ]. If 2

0

1 ( ( ))′−∫x

f t 0

( ) , 0 1,= ≤ ≤∫x

dt f t dt x and

(0) 0,=f then

a. 1 1

2 2

<

f and 1 1

3 3

>

f

b. 1 1

2 2

>

f and 1 1

3 3

>

f

c. 1 1

2 2

<

f and 1 1

3 3

<

f

d. 1 1

2 2

>

f and 1 1

3 3

<

f

10. Let α and β be the roots of 2 6 2 0,x x− − = with .α β> If

n n

na α β= − for 1,n ≥ then the value of 10 8

9

2

2

a a

a

−is

a. 1 b. 2

c. 3 d. 4

11. Let cos sin .θ θ= +z i Then the value of 15

2 1

1

Im( )−

=∑ m

m

z at

2θ = ° is

a. 1

sin 2° b.

1

3 sin 2°

c. 1

2 sin 2° d.

1

4 sin 2°

12. If is a matrix satisfying the equation

where I is identity matrix, then the

ordered pair (a, b) is equal to:

a. b.

c. d.

13. The sum of first 20 terms of the sequence

is

a. b.

c. d.

9x

2 3 100(1 ) (1 ) (1 ).....(1 )x x x x+ + + +

[0, 2 ]π

250πV

24=y x 4 0.+ + =x y

5,= −y

:f R R→

(1) .2

f1

=

1( ) ( )

x

F x f t dt−

= ∫ [1 1, 2]x∈ −1

( ) | ( ( )) |x

G x t f f t dt−

= ∫

[ 1, 2].x∈ −1

( ) 1lim ,

( ) 14x

F x

G x→= 1

2f

ˆˆ ˆ ; , , 1,1.V ai bj ck a b c= + + ∈ −

2pV in

1 2 2

2 1 2

2

= −

A

a b

9 ,=TAA I 3 3×

(2, 1)− ( 2,1)−

(2,1) ( 2, 1)− −

0.7, 0.77, 0.777, .....

207(179 10 )

81

−− 207(99 10 )

9

−−

207(179 10 )

81

−+ 207(99 10 )

9

−+

Mathematics158

14. Coefficient of in the expansion of

is

a. 1051 b. 1106

c. 1113 d. 1120

15. The number of seven digit integers, with sum of the digits

equal to 10 and formed by using the digits 1, 2 and 3 only,

is

a. 55 b. 66

c. 77 d. 88

16. The mean of the data set comprising of 16 observations is

16. If one of the observation valued 16 is deleted and three

new observations values 3, 4 and 5 are added to the data,

then the mean of the resultant data, is

a. 16.8 b. 16.0

c. 15.8 d. 14.0

17. Let ABC be a triangle such that 6

ACBπ

∠ = and let a, b

and c denote the lengths of the sides opposite to A, B and

C respectively. The value (s) of x for which

2 21, 1a x x b x= + + = − and 2 1c x= + is (are)

a. (2 3)− + b. 1 3+

c. 2 3+ d. 4 3

18. Let where and

Then equals:

a.

b.

c. d.



Column II.

19. Match the Column:

Column I Column II

(A) In if the magnitude

of the projection vector

of the vector

on is and

if then

possible value(s) of

is (are)

1. 1

(B) Let a and b be real

numbers such that the

function

is

differentiable for all

Then possible

value(s) of a is (are)

2. 2

(C) Let be a

complex cube root of

unity. If

then possible value(s)

of n is (are)

3. 3

(D) Let the harmonic mean

of two positive real

numbers a and b be 4. If

q is a positive real

number such that a, 5,

q, b is an arithmetic

progression, then the

value(s) of is

(are)

4. 4

5. 5

20. Consider the following linear equations

0+ + =ax by cz

0+ + =bx cy az

0+ + =cx ay bz

Match the conditions / expressions in Column I with

statements in Column II

Column I Column II

(A) 0+ + ≠a b c and

2 2 2+ + = + +a b c ab bc ca

1. the equations represent

planes meeting only at

a single point.

(B) 0+ + =a b c and

2 2 2a b c ab bc ca+ + ≠ + +


the line .= =x y z

(C) 0+ + ≠a b c and 2 2 2

a b c ab bc ca+ + ≠ + +


identical planes.

(D) 0+ + =a b c and 2 2 2+ + = + +a b c ab bc ca


the whole of the three

dimensional space.

11x

2 4 3 7 4 12(1 ) (1 ) (1 )+ + +x x

1( ) (sin cos ),

k k

kf x x xk

= + x R∈ 1.k ≥

4 6( ) ( )f x f x−

1

6

1

3

1

4

1

12

2 ,R

ˆ ˆα β+i j

ˆ ˆ3 +i j 3

α | 2 3β,= +

|α |

2

2

3 2, 1( )

, 1

− − <=+ ≥ax x

f xbx a x

.∈x R

ω 1≠

2 4 3(3 3 2 ) +− +ω ω n

2 4 3(2 3 3 ) ++ + −ω ω n

2 4 3( 3 2 3 ) 0,

++ − + + =ω ω n

| |−q a


159Mock Test-2




9 (both inclusive).

1. The number of distinct real roots of

4 3 24 12 1 0− + + − =x x x x is

2. The number of all possible values of ,θ where 0 ,θ π< <

for which the system of equations

( )cos3 ( )sin 3y z xyzθ θ+ =2cos3 2sin3

sin3xy z

θ θθ = +

( )sin ( 2 )cos3 sin3xyz y z yθ θ θ3 = + + have a solution

0 0 0( , , )x y z with 0 0 0,y z ≠ is

3. Let , Rα β ∈ be such that 2

0

sin( )lim 1.

sinx

x x

x x

βα→

=−

Then

6( )α β+ equals

4. The centres of two circles and each of unit radius

are at a distance of 6 units from each other. Let P be the

mid-point of the line segment joining the centres of

and and C be a circle touching circles and

externally. If a common tangent to and C passing

through P is also a common tangent to and C, then the

radius of the circle C is

5. Suppose that and are three non-coplanar vectors in

Let the components of a vector along and be

4, 3 and 5, respectively. If the components of this vector

along and are x, y

and z, respectively, then the value of is

6. Suppose that the foci of the ellipse are

and where and Let and

be two parabolas with a common vertex at (0, 0) and

with foci at and respectively. Let be

a tangent to which passes through and be

a tangent to which passes through The is

the slope of and is the slope of then the value

of is

7. Let f :IR IR be defined as f(x) = The total

number of points at which f attains either a local

maximum or a local minimum is

8. The total number of distinct x R∈ for which

2 3

2 3

2 3

1

2 4 1 8 0

3 9 1 27

x x x

x x x

x x x

+

+ =

+

is



9. If and

where N is the set of natural numbers, then is

equal to:

a. N

b. Y – X c. X d. Y

10. A value of b for which the equations2

2

x bx 1 0x x b 0,

+ − =+ + =

Have one root in common is

a. 2− b. i 3− c. i 5 d. 2

11. The variance of first 50 even natural numbers is:

a.

b. 833 c. 437 d.

12. The expression can be written as

a. b. sec cosec 1A A +

c. d. sec cosecA A+

13. The value of

a. b.

c. d.

14. A man is walking towards a vertical pillar in a straight

path, at a uniform speed. At a certain point A on the path,

he observes that the angle of elevation of the top of the

pillar is After walking for 10 minutes from A in the

same direction, at a point R, he observes that the angle of

elevation of the top of the pillar is Then the time

taken (in minutes) by him, from B to reach the pillar, is:

a. 6 b. 10

c. 20 d. 5

1C 2C

1C

2C 1C 2C

1C

2C

,p qr r

rr

3.R sr

,p qr r

rr

sr

( ),p q r− + +r r r

( )p q r− +r r r

( )p q r− − +r r r

2x y z+ +

2 2

19 5

x y+ =

1( , 0)f 2( , 0)f 1 0f > 2 0.f < 1P

2P

1( , 0)f 2(2 , 0),f 1T

1P 2(2 , 0)f 2T

2P 1( , 0).f 1m

1T 2m 2 ,T

2

22

1m

m

+

→ 2| | | 1 | .x x+ −

4 3 1: nX n n N= − − ∈ 9( 1) : ,Y n n N= − ∈

X Y∪

833

4

437

4

tan cot

1 cot 1 tan

A A

A A+

− −

sin cos 1A A +

tan cotA A+

231

1 1

cot cot 1 2 isn

n k

k−

= =

+

∑ ∑

23

25

25

23

23

24

24

23

30 .°

60 .°

Mathematics160

15. is equal to

a. b. c. 1 d. 2

16. Let the function be given by

Then, is

a. even and is strictly increasing in

b. odd and is strictly decreasing in

c. odd and is strictly increasing in

d. neither even nor odd, but is strictly increasing in






A box contains 1 white ball, 3 red balls and 2 black balls.

Another box contains 2 white balls, 3 red balls and 4 black

balls. A third box B3 contains 3 white balls, 4 red balls and 5

black balls.

17. If 1 ball is drawn from each of the boxes and

the probability that all 3 drawn balls are of the same

colour is

a. b. c. d.

18. If 2 balls are drawn (without replacement) from a

randomly selected box and one of the balls is white and

the other ball is red, the probability that these 2 balls are

drawn from box is

a. b. c. d.


Suppose we define the definite integral using the following

formula for more accurate

result for

When

19. is equal to

a. b.

c. d.

20. If and c is a point such that

and is the point lying on the curve for

which is maximum, then is equal to

a. b.

c. d. 0

0

(1 cos2 ) (3 cos )lim

tan 4x

x x

x x→

− +

1

4−

1

2

: ( , ) ,2 2

π π −∞ ∞ → −

g

1( ) 2 tan ( ) .2

π−= −ug u e

g

(0, )∞

( , )−∞ ∞

( , )−∞ ∞

( , )−∞ ∞

1B

2B

1 2,B B 3 ,B

82

648

90

648

558

648

566

648

2B

116

181

126

181

65

181

55

181

( )2

b

a

d af x dx

−=∫ ( ( ) ( )),f a f b+

( , ) ( ) ( ( ) ( ))2

c ac a b F c f a f c

−∈ = + +

2

b c−

( ( ) ( )).f b f c+ ,2

a bc

+=

( ) ( ( ) ( ) 2 ( )).4

b

a

b af x dx f a f b f c

−= + +∫

/ 2

0

sin x dx

π

∫

(1 2)8

π+ (1 2)

4

π+

8 2

π4 2

π

( ) 0 ( , )f x x a b′′ < ∀ ∈

,a c b< < ( , ( ))c f c

( )F c ( )f c′

( ) ( )f b f a

b a

−−

2( ( ) ( ))f b f a

b a

−−

2 ( ) ( )

2

f b f a

b a

−−


161Mock Test-2

ANSWER & SOLUTIONS

JEE-Main

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

a d c d b a d d b b

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

a c a b c a a a b b

21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

b d b a d c a d a b

1. (a) Use the identity

replace x by in the given function we get

solve to find

2. (d) Given

But and

Hence the equation will be same.

3. (c)

and

∴

4. (d)

5. (d) It is understandable.

6. (a) Given series is

(in G.P.)

Here then sum of the five terms

7. (d) .

Putting and and adding the results

8. (d)

9. (b) The given series reduces to

10. (b) corresponds to none of the lamps is

being switched on.

11. (a) The event that the fifth toss results in a head is

independent the event that the first four tosses result in tails.

Probability of the required event = ½

12. (c) .

13. (a)

. . .(i)

Now,

( 1( ))f f x x− =

1( ),f x−

1

1( 1( )) 1( )

( )f f x f x

f x−− = − +

⇒1

11( ) ,

( )x f x

f x−= − + 1 .f x−

2 1 0x x+ + =

∴ 1 1 1[ 1 3] ( 1 3), ( 1 3)

2 2 2x i i i= − ± = − + − − 2,ω ω=

19 19α ω ω= = 7 14 2 .β ω ω= =

2 2cos sinn i

n n

π πω = +

⇒ 3

2 2 1 3cos sin

3 3 2 2

ii

π πω ω= + = − + =

2

2

3

2 2 4 4cos sin cos sin

3 3 3 3i i

π π π πω = + = +

21 3.

2 2

iω= − − =

2 2

3 3 3 3( ) ( )x y z x y zω ω ω ω+ + + +2 2( ) ( )x y z x y zω ω ω ω= + + + +

2 2 2 .x y z xy yz zx= + + − − −

6

1

2 2sin cos

7 7k

k ki

π π

=

−

∑

6

1

2 2cos sin

7 7k

k ki i

π π

=

= − +

∑2

6 7

1

i k

k

i e

π

=

= − ∑

2 / 7 4 / 7 6 / 7 8 / 7 10 / 7 12 / 7 i i i i i ii e e e e e eπ π π π π π= − + + + + +

12 / 72 / 7

2 / 7

(1 )

1

ii

i

ei e

e

ππ

π

−= −

−

2 / 7 14 / 7

2 / 7

)

1

i i

i

e ei

e

π π

π

−= −

−

14 / 7( 1)ie π =Q

2 / 7

2 / 7

1

1

i

i

ei i

e

π

π

−= − =

−

1 3 273 4 6 ........ 3 .....

2 4 2 4

9+ + + = + + +

2 3 4 53 3 3 33 .....

2 4 8 16= + + + + +

33, ,

2a r= =

55

5

3 33 1 1 1

2 32( 1)

3 111

2 2

na rS

r

− − − = = =− −

243 32 211 3 633 96 39 .

32 16 16 16

− × = = = =

2 6 2 12

1 2 12(1 2 ) 1 ....x x a x a x a x+ − = + + + +

1x = 1x = −

2 464 2(1 ...)a a= + + +

∴2 4 6 12.... 31.a a a a+ + + + =

( 1)

! 2( )!n

n n nT

n n

Σ += =

1 ( 1) 1 1 2

2 ( 1)! 2 ( 1)! ( 1)!

n n

n n n

+ −= = + − − −

1 1 2 ( 2 ) 3.

2 ( 2)! ( 1)! 2 2

e e e

n n

+= + = = − −

3 4log 2 log log log

2 3 1e e e e

n

n

+ + + + − K

log 2 log 3 log 2 log 4 log 3e e e e e= + − + − +K

log ( ) log ( 1)e en n+ − − log .e n=

102 1 1023,− =

∴

1 1 3.2. . 2 8

42 2.3 4 2

xx x= ⇒ = ⇒ =

sin sin3 sin 2 sinθ θ θ α+ + =

⇒ 2sin 2 cos sin 2 sinθ θ θ α+ =⇒ sin 2 (2 cos 1) sinθ θ α+ =

cos cos3 cos 2 cosθ θ θ α+ + =2cos 2 cos cos 2 cosθ θ θ α+ =

Mathematics162

. . .(ii)

From (i) and (ii),

⇒ ⇒ .

14. (b) Imaginary part of

15. (c)

cot 30 cot 60d h h= ° − ° and time = 3 min.

∴ Speed (cot 30 cot 60 )

3

o oh −= per minute

It will travel distance cot 60h ° in

cot 60 3

1.5(cot 30 cot 60 )

o

o o

h

h

×=

− minute.

16. (a) For continuous

⇒

17. (a) x yy e += ⇒ log ( ) logy x y e= +

⇒ 1

1dy dy

y dx dx= + ⇒ .

1

dy y

dx y=

−

18. (a) Let

Put

Point of maxima

19. (b) Put then

20. (b)

Required area

sq. unit.

21. (b) ⇒

On integrating, we get

22. (d)

⇒

Put and

23. (b)

2 2

2 2 1 1| |

51 2AD

− −= =

+

⇒ tan 60AD

BD° =

⇒ 1/ 5

3BD

=

⇒ 1

15BD =

⇒ 2 2 / 15BC BD= = .

⇒ cos 2 (2cos 1) cosθ θ α+ =

⇒ tan 2 tanθ α= 2θ α= / 2θ α=

1 5 7 9sin

16 16

i−

−

9 9log 1 log(2).

16 16

= − + + = −

2lim ( ) (2)x

f x f k→

= =

3 2

22

16 20lim

( 2)x

x x xk

x→

+ − +=

−2

22

( 4 4) ( 5)lim 7.

( 2)x

x x x

x→

− + += =

−

sin .cosp qy x x=

1 1sin .cos .cos cos .( sin )sinp q q pdyp x x x q x x x

dx

− −= + −

1 1 1 1sin .cos cos .sinp q q pdyp x x q x x

dx

− + − += −

0,dy

dx=

2tan

px

q∴ =

⇒ tanp

xq

= ±

∴ 1tan .p

xq

−=

1log ,x t dx dt

x= ⇒ =

2 2

1 1 1 1.

(log ) logdx dt c c

x x t t x= = − + = − +∫ ∫

' ' 2CDD C ABCD= = ×

41/ 2

1

28 22 2

3x dx= =∫

210

dy x

dx x

++ =

1 0dy x dx

x

+ + =

2

log 0.2

xy x c+ + + =

cos log(sec tan ) cos log(sec tan )y x x dx x y y dy+ = +

sec log(sec tan )y y y dy+∫sec log(sec tan )x x x dx= +∫log(sec tan )x x t+ = log(sec tan )y y z+ =

2 2[log(sec tan )] [log(sec tan )].

2 2

x x y yc

+ += +

60o

A (2, –1)

C D

x + 2y –1 = 0

xy 2=

x =1 x = 4

D C

X B A

D′ O

Y

C′

30° 60°

h

d

163Mock Test-2

24. (a) It represents a circle, if

.

25. (d) Tangent to the curve is

So, it must satisfy

Since, it has equal roots.

z

Hence, equation of common tangent is

26. (c) Since and are

collinear, therefore

or

are linearly independent)

27. (a)

and

∴

Therefore,

28. (d)

29. (a)Plane is , where

or . . .(i)

Now according to equation,

Put the values of x, y, z in (i), we get the locus of the

centroid of the tetrahedron.

30. (a) .

JEE Advance Paper-I

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

8 8 4 8 4 4 7 5 c c

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

d d c c c d b d c a

1. (8) can be formed in 8 ways

i.e.,

and coefficient in each case is 1

Coefficient of

2. (8) ( )5 4 3 8 10

1/85 4 3 3 8 103( ) 1

( ) (1) ( )( ) 11 1 3 1 1 1

a a a a aa a a a a

− − −− − −+ + + + +

≥ =+ + + + +

(using AM GM)≥

∴ 5 4 3 8 103 1 8− − −+ + + + + ≥a a a a a

3. (4) 3/ 2

1 1 1 16 log 4 4 4 ...

3 2 3 2 3 2 3 2+ − − −

1 1

put 4 4 ...3 2 3 2

x = − −

2 4

3 2

xx = −

23 2 12 2x x= −

23 2 12 2 0x x+ − =

⇒ 4 2

; 3/ 23

x x= = − not possible

⇒ 3/ 2

1 4 26 log

33 2

+ ×

⇒ 3/ 2

46 log

9

+

⇒ 6 2 4− =

4. (8)

Number of solutions = 8

5. (4) Let inner radius be r and inner length be

Volume of material be M

a b=

⇒1

3 4

k= ⇒

3

4k =

2 8y x= 2 / .y mx m= +

1.xy = −

⇒ 22 21 1 0x mx mx x

m m

+ = − ⇒ + + =

∴ 0D =

⇒ 3

2

44 0 1m m

m− = ⇒ =

⇒ 1m = 2.y x= +

( 2)c x a b= − +rr r

(2 1)d x a b= + −r rr

c dλ=rr

⇒ ( 2) (2 1)x a b x a bλ λ− + = + −r rr r

[( 2) (2 1)] ( 1) 0x x a bλ λ− − + + + =rr

( 2) (2 1) 0, 1 0x xλ λ− − + = + =

( ,a brr

Q

⇒ 2 2 1 0x x− + + =

⇒1

.3

x =

( ) ( )2 2 2

a b a b ab a c b c b a c a b

+ +− ⋅ − = ⋅ − ⋅ − ⋅ + +

r rr r rr r r rr r r r r r

| | | |a c b c− = −rr r r

⇒ 2 2| | | |a c b c− = −rr r r

2a b c+ =rr r

( ) 02

a bb a c

+− ⋅ − =

rrr r r

1 1(2 10 12)cos cos (0)

4 25 16 1 4 9θ − −+ −

= =+ + + +

⇒ 90θ = °

1x y z

a b c+ + =

2

1

(1/ )p

a=

Σ

2 2 2 2

1 1 1 1

a b c p+ + =

, ,4 4 4

a b cx y z= = =

qpqp ~~)(~ ∧≡∨

9x

9 1 8 2 7 3 6 4 5 1 2 6 1 3 5 2 3 4, , , , , , ,+ + + + + + + + + +x x x x x x x x

⇒ 9

8 times1 1 1 ......... 1 8= + + + + =x

2 4 4 6 65cos 2 cos sin cos sin 2

4x x x x x+ + + + =

⇒2 2 25

cos 2 5cos sin 04

x x x− =

⇒2tan 2 1, where 2 [0, 4 ]x x π= ∈

l

2r Vπ =l

2 2( 2) ( 2)M r rπ π= + + −l l

2 3

4 88 0 4

dM V Vr

dr r rπ π= − − + + +

Mathematics164

⇒

⇒

6. (4) Image of

about the line is

7. (7)

⇒

Given

⇒

8. (5) Let be vectors

rest of the vectors are and let us

find the number of ways of selecting co-planar vectors.

Observe that out of any 3 coplanar vectors two will be

collinear (anti parallel)

Number of ways of selecting the anti parallel pair = 4

Number of ways of selecting the third vector = 6

Total = 24

Number of non co-planar selections

Alternate

Required value

9. (c)2

1′ = ± −f f

⇒ ( ) sin=f x x or ( ) sin= −f x x (not possible)

⇒ ( ) sin=f x x

Also, sin 0> ∀ >x x x

10. (c)α is a roots of equation

2 26 2 0; 6 2 0Bα α β− − = − − =

2 6 2 0α α− − =

⇒ 2 2 6α α− =

⇒ 10 10 8 8

10 8

9 9

9

2 ( ) 2( )

2 2( )

a a

a

α β α βα β

− − − −=

−

8 2 8 2

9 9

( 2) ( 2)

2( )

α α β βα β

− − −=

−

8 8

9 9

.(6 ) (6 )

2( )

α α β βα β

−=

−

9 9

9 9

6 ( )3

2 2( )

α βα β

−− =

−

11. (d) sin sin3 ... sin 29θ θ θ= + + +X

2(sin ) 1 cos2 cos2 cos4 ... cos28 cos30θ θ θ θ θ θ= − + − + + −X

⇒ 1 cos30 1

2sin 4sin 2

θθ

−= =

°X

12. (d)

Equation

. . .(i)

and

. . .(ii)

. . .(iii)

Solving

13. (c)

0 when 10dM

rdr

= =

1000V π=

4250

V

π=

5y = −

4 0x y+ + = 1x =

⇒ Distance 4AB =

1

1(1) | ( ( )) | 0

−= =∫G t f f t dt

( ) ( )− = −f x f x

1(1)

2=f

1 1

( ) (1)

( ) (1) 11lim lim( ) (1)( ) | ( (1)) | 14

1

→ →

−−= = =−−

x x

F x F

F x fxG x GG x f f

x

⇒1/ 2 1

| (1/ 2) | 14=

f⇒

17.

2

=

f

(1,1,1), ( 1,1,1), (1, 1,1), ( 1, 1, 1)− − − − −

, , ,a b c dr rr r

, , ,a b c d− − − −r rr r

8 5

3 24 32 2 , 5C p= − = = =

8 6 4

3!

× ×=

∴ 5p =

9=TAA I

1 2 2 1 2

2 1 2 2 1 2 9

2 2 2

− = −

a

I

a b b

⇒2 2

9 0 4 2 9 0 0

0 9 2 2 2 0 9 0

4 2 2 2 2 4 0 0 9

a b

a b

a b a b a b

+ + + − ⇒ + + + − + +

4 2 0+ + =a b

⇒ 2 4a b+ = −

2 2 2 0a b+ − =

⇒ 2 2 2a b− = −

2 24 0a b+ + =

⇒ 2 2 5a b+ =

2, 1a b= − = −

0.7 0.77 0.777 ..... 0.777...7+ + + +

7[0.9 0.99 0.999 ... 0.999...9]

9= + + + +

7[(1 0.1) (1 0.01) (1 0.001...1) ... (1 0.000...1)]

9= − + − + − + + −

2 3 20

7 1 1 1 120 ...

9 10 10 10 10

= − + + + +

2020

20

11

7 1 7 1 10 11020 . 20 .19 10 9 9 10

110

− −= − = −

−

20

20

7 1 7180 1 [179 10 ]

81 10 81

− − − = +

165Mock Test-2

14. (c)

Possibilities are (0, 1, 2); (1, 3, 0); (2, 1, 1); (4, 1, 0).

Required coefficients

15. (c) Coefficient of 10x in 2 3 7( )+ +x x x

Coefficient of 3x in 2 7(1 )+ +x x

Coefficient of 3x in

3 7 7(1 ) (1 )−− −x x 7 3 7

3 7+ += −C 9

3 7= −C

9 8 77 77

6

× ×= − =

16. (d)

17. (b) Using cosine rule for C∠

2 2 2 2 2

2 2

3 ( 1) ( 1) (2 1)

2 2( 1)( 1)

x x x x

x x x

+ + + − − +=

+ + −

⇒

2

2

2 2 13

1

x x

x x

+ −=

+ +

⇒ 2( 3 2) ( 3 2) ( 3 1) 0x x− + − + + =

⇒ (2 3) 3

2( 3 2)x

− ±=

−

⇒ (2 3),1 3 1 3x x= − + + ⇒ = + as ( 0).x >

18. (d)

19. (A)

. . .(i)

Given . . .(ii)

From equation (i) and (ii), we get

So

(B)

For continuity

⇒ . . .(i)

For differentiability

⇒

⇒

⇒

(C)

(D) Let

Given

⇒

⇒

⇒

⇒

⇒

⇒

20.

a b c

b c a

c a b

∆ = 2 2 21( )[( ) ( ) ( ) ]

2a b c a b b c c a= − + + − + − + −

(A) If 0+ + ≠a b c

and 2 2 2+ + = + +a b c ab bc ca

⇒ 0∆ = and 0= = ≠a b c

⇒ the equations represent identical planes.

(B) 0+ + =a b c and 2 2 2+ + ≠ + +a b c ab bc ca

⇒ 0∆ =

⇒ the equations have infinitely many solutions.

( )+ + +ax by a b z

( )+ = +bx cy b c z

⇒ 2 2( ) ( )− = −b ac y b ac z ⇒ =y z

⇒ 0+ + =ax by cy ⇒ =ax ay ⇒ = =x y z

1 2 32 3 4 11x x x+ + =

∴4 7 12 4 7 12 4 7 12 4 7

0 1 2 1 3 0 2 1 1 4 1( ) ( ) ( ) ( 1)C C C C C C C C C C C= × × + × × + × × + × ×

(1 7 66) (4 35 1) (6 7 12) (1 7)= × × + × × + × × + ×

462 140 504 4 113.= + + + =

1616

ixΣ= ⇒ 256ixΣ =

( ) 16 3 4 5 25214

18 18

ixΣ − + + +

= =

1(sin cos )

4

k k

kf x x= + 6 6

6

1( ) (sin cos )

6f x x x= +

4 4

4

1( ) (sin cos )

4f x x x= +

2

6

1 31 sin 2

6 4f K x

= − 2

4

1 sin 2( ) 1

4 2

xf x

= =

2 2

4 6

1 sin 2 1 sin 2 1 1 1( ) ( )

4 8 6 8 4 6 12

x xf x f x

− = − − − = − =

33

2

α β+=

3 2 3α β+ = ±

2 3α β= +

α 2 1or= −

|α| 1 or 2=

2

2

13 2,( )

1,

xaxf x

xbx a

<− −=

≥+2

3 2a b a− − = +2

3 2a a b+ + = −

6a b− =

6a b= −2

3 2 0a a− + =1,2a =

2 4 3 2 4 3 2 4 3(3 3 2 ) (2 3 3 ) ( 3 2 3 ) 0

n n nω ω ω ω ω ω+ + +− + + + − + − + + =2 4 3 2 4 3 2 2 4 3(3 3 2 ) ( (2 3 3 )) ( ( 3 2 3) 0n n nω ω ω ω ω ω ω ω+ + +− + + + − + − + + =

⇒ 2 4 3 4 8(3 3 2 ) (1 ) 0n n nω ω ω ω+− + + + + =

⇒ 3 ,n k k N≠ ∈

5a d= −

5q d= +

5 2b d= +

| | | 2 |q a d− =

24

ab

a b=

+

2ab

a b=

+

(5 )(5 2 ) 2(5 5 2 ) 2(10 )d d d d d− + = − + + = +2

25 10 5 2 20 2d d d d+ − − = +2

2 3 5 0d d− − =

51,

2d d= − =

| 2 | 2,5d =

Mathematics166

(C) 0+ + ≠a b c

and 2 2 2+ + ≠ + +a b c ab bc ca

⇒ 0∆ ≠

⇒ the equation represent planes meeting at only one point.

(D) 0+ + =a b c

and 2 2 2+ + = + +a b c ab bc ca

⇒ 0= = =a b c

⇒ the equation represent whole of the three dimensional space.


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

2 3 7 8 9 4 5 2 d b

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

b b b d d c a d a a

1. (2) Let 4 3 2( ) 4 12 1 0= − + + − =f x x x x x

3 2 3 2( ) 4 12 24 1 4( 3 6 ) 1′ = − + + = − + +f x x x x x x x

2 2( ) 12 24 24 12( 2 2)′′ = − + = − +f x x x x x

( )′′f x has 0 rl roots f(x)has maximum 2 distinct real roots

as (0) 1.= −f

2. (3) ( )cos3 ( )sin 3 0y z xyzθ θ+ − = . . .(i)

sin 3 (2cos3 ) (2sin 3 )xyz z yθ θ θ= + . . .(ii)

∴ ( ) cos3 (2cos3 ) (2sin 3 ) ( 2 )cos3 sin 3y z z y y z y yθ θ θ θ θ+ = + = + +

(cos3 2sin 3 ) cos3zθ θ θ− = and (sin cos3 )y θ θ− = 0

⇒ 0 sin3 cos3 0θ θ⇒ − = ⇒ ⇒ sin 3 cos3θ θ=

∴ 3 / 4nθ π π= +

3. (7)

3 3 5 52

2

3 50 0

......3! 5!sin

lim limsin

.......3! 5!

x x

x xx x

x x

x x x xx x

β ββ

βα

α→ →

− + −

=−

− − − −

3 23

3 50

.....3!

lim 1

( 1) .......3! 5!

x

xx

x xx

ββ

α→

− +

= =− + − +

⇒ 1 0α − =

⇒ 1,α =

Limit 6 1β= =

⇒ 1

6β =

⇒ 1 7

6( ) 6 1 6 76 6

α β + = + = × =

4. (8)

5. (9)

On solving we get

6. (4) The equation of is and is

Tangent to passes through (–4, 0)

⇒

Also tangent to passes through (2, 0)

⇒

7. (5)

So, total no. of local maxima and local minima is = 5

2 2cos

3α =

1sin

3α =

2 2tanα =

R

⇒2 2

8units.tanα

= =R

4 3 5= + +r r r rs p q r

( ) ( ) ( )= − + + + − + + − − +r r r r r r r r r rs x p q r y p q r z p q r

( ) ( ) ( )= − + − + − − + + +r r r rs x y z p x y z q x y z r

⇒ 4− + − =x y z ⇒ 3− − =x y z ⇒ 5+ + =x y z

9 74, ,

2 2= = = −x y z

⇒ 2 9+ + =x y z

1P2y 8x 0− =

2P

2y 16x 0+ =2y 5x 0− =

1

1

20 m ( 4)

m= − +

⇒2

1

12

m=

2y 16x 0+ =

2

2

40 m 2

m= × −

⇒ 2

2m 2=

⇒ 2

22

1

1m 4

m+ =

2 1fx x x= + −

2

2

2

2

if 11

if –1 x<01

if 0 1–x 1

if 1x –1

xx x

x x

xx

xx

≤ − − − ≤− − +

= ≤ <+ +

≥+

1C

A2 2

P αα

3

2C

1

B

C

R α

167Mock Test-2

8. (2)

3

2 3

3

1 1 1

0 2 6 1 0

0 6 24 2

x

x x x

x

+

⋅ − =

−

⇒ 3 3(12 2) 0x x + =

⇒ 6 36 5 0x x+ − =

⇒ 6 3 36 6 5 5 0x x x+ − − =

⇒ 3 3(6 5)( 1) 0x x− + =

⇒ 3 3 5

1,6

x x= − =

⇒

1/35

1,6

x x

= − =

So, two solution

9. (d)

⇒

10. (b) 2x bx 1 0+ − =

2x x b 0+ + = . . .(i)

Common root is (b 1)x 1 b 0− − − =

⇒ b 1

xb 1

+=

− This value of x satisfies equation (i)

⇒

2

2

(b 1) b 1b 0

b 1(b 1)

+ ++ + =

−−

⇒ b 3i, 3i, 0.= −

11. (b) Variance =

12. (b) Exp.

13. (b)

14. (d) tan

for 2y distance time

= 10 min.

So for y dist time = 5 min.

15. (d)

16. (c)

is odd and

increasing.

17. (a) P (required) = P (all are white) + P (all are red) + P

(all are black)

18. (d) Let A : one ball is white and other is red

Both balls are from box


4 3 1,n

x n n N= − − ∈

(1 3) 3 1,nx n n N= + − − ∈

⇒ 0,9,54, ...X = 9( 1),y n n N= − ∈

0,9,18, ...y = ⇒ .x y y∪ =

( )2

21

xx

n−∑

⇒22 2 2 2

2 2 4 6 ... 100 2 4 ... 100

50 50σ

+ + + + + + + −

⇒ 23434 2601 833σ = − =

22

2

tan 1 1 1tan

tan 1 tan tan tan 1 tan

AA

A A A A A

= + = − − − −

2tan tan 1

tan cot 1tan

A AA A

A

+ += = + + sec . 1A cosec A= +

231 2

1

cot cot ( 1)n

n n−

=

+ +

∑

231

1

1cot tan

1 ( 1)n

n n

n n

−

=

+ − + + ∑

⇒ 1 23 25cot tan .

325 23

− =

130

3° = =

+x

y z

⇒ 3 = +x y z

⇒ tan 60 3° = =x

y

⇒ 3= = +x y y z

3 = +y y z

⇒ 2 =y z

0

(1 cos2 ) (3 cos )lim

tan 4x

x x

x x→

− +

2

0

(2sin )(3 cos )lim

tan 44

4

x

x x

xx x

x

→

+=

×

2

20

2sin (3 cos ) 2lim (3 1) 2

4 4x

x x

x→

+= = + =

1( ) 2 tan ( )2

ug u eπ−= −

1 1 1 1 12 tan tan cot tan cot− − − − −= − − = −u u u u ue e e e e

( ) ( )− = −g x g x

⇒ ( )g x ( ) 0′ >g x

⇒

1 2 3 3 3 4 2 4 5 5

6 9 12 6 9 12 6 9 12 12= × × + × × + × × ×

6 36 40 82.

648 648 648 648= + + =

1:E

1B

2:E

2B

y B z A

x

60 30

Mathematics168


Here, P (required)

19. (a)

20. (a)

⇒

3:E

3B

2 =

EP

A

2

2

1 2 .3

1 2 3

. ( )

A. ( ) . ( ) . ( )

=

+ +

AP P E

E

A AP P E P P E P E

E E E

2 3

1 1

9

2

1 3 2 3 3 4

1 1 1 1 1 1

6 9 12

2 2 2

1 1

3 5561 1 2 181.1 1 1

5 6 113 3 3

××

= = =× × × + +× + × + ×

C C

C

C C C C C C

C C C

/ 2

0

0 02 2sin sin(0) sin 2sin

4 2 2x dx

ππ π

π + + = + +

∫

(1 2)8

π= +

( ) ( ) ( ) ( ) ( )F c b a f c f a f b′ ′= − + −

( ) ( )( ) 0F c f c b a′′ ′′= − <

⇒ ( ) 0F c′ =

( ) ( )( )

f b f af c

b a

−′ =

−

169Mock Test-3

JEE-MAIN: MATHEMATICS MOCK TEST -3

1. Let and Then for all

is equal to:

a. x b. 1 c. d.

2. If then

a. x is an irrational number b.

c. d. None of these

3. If then is equal to

a. i b. – I c. 1 d. – 1

4. The inverse of is

a. b.

c. d.

5. The sum of 100 terms of the series ill be

a. b.

c. d.

6. If n is an integer greater than 1, then

a. b. 0

c. d.

7. The sum of is

a. b.

c. d.

8. If the letters of the word KRISNA are arranged in all

possible ways and these words are written out as in a

dictionary, then the rank of the word KRISNA is

a. 324 b. 341


9. Seven white balls and three black balls are randomly

placed in a row. The probability that no two black balls

are placed adjacently, equals

a. b.

c. d.

10. If , then =

a. b.

c. d.

11. Find real part of

a. b.

c. d. None of these

12. A house of height 100 metres subtends a right angle at the

window of an opposite house. If the height of the window

be 64 metres, then the distance between the two houses is

a. 48 m b. 36 m c. 54 m d. 72 m

13. If then

a. b.

c. is continuous at d. None of these

14. If ,y x yx e −= then dy

dx=

a. 2log .[log( )]x ex −

b. 2log .[log( )]x ex

c. 2log .(log )x x d. None of these

15. 20 is divided into two parts so that product of cube of one

quantity and square of the other quantity is maximum. The

parts are

a. 10, 10 b. 16, 4

c. 8, 12 d. 12, 8

16.

a. b.

c. d. None of these

( ) 1 [ ]g x x x= + −

1 0

( ) 0 0.

1 1

x

f x x

x

− <

= = >

, ( )x f g x

( )f x ( )g x

6 6 6 ....to ,x = + + + ∞

2 3x< <

3x =

1 1,z z−+ = 100 100z z−+

2 3

4 2

− −

2 31

4 28

−

3 21

2 48

−

2 31

4 28

3 21

2 48

.9 .09 .009.........+ +100

11

10

−

1001

110

+

1061

110

−

1001

110

+

1( 1)n na C a− − +

2( 2)C a − .... ( 1)n+ + − ( )a n− =

a

2a 2n

2 6 12 20

1! 2! 3! 4!+ + + +K

3

2

ee

2e 3e

1

2

7

15

2

15

1

3

3

2

ππ α< <

1 cos 1 cos

1 cos 1 cos

α αα α

− ++

+ −

2

sinα2

sinα−

1

sinα1

sinα−

1 3cos

2 2

i− +

/3π / 4π

3 1log

2

−

sincos , when 0

( )

2, when 0

xx x

f x x

x

+ ≠= =

0lim ( ) 2x

f x→ +

≠0

lim ( ) 0x

f x→ −

=

( )f x 0x =

4 21tan secx x dx

x=∫

52 tan x c+ 51tan

5x c+

52tan

5x c+

Mathematics170

17. The area bounded by the straight lines and the

curves is

a. b.

c. d.

18. The solution of is

a.

b.

c.

d.

19. The product of the perpendiculars drawn from the points

on the line , is

a. b.

c. d.

20. Area of the circle in which a chord of length 2 makes

an angle 2

πat the centre is

a. 2

π b. 2π

c. π d. 4

π

21. The focal chord to is tangent to

then the possible values of the slope of

this chord, are

a. b.

c. d.

22. If are non-zero vectors such that then

which statement is true

a. b.

c. or d. None of these

23. The point of intersection of the lines

is

a. b.

c. d.

24. is equal to

a. b.

c. d.

25. The value of b and c for which the identity

is satisfied, where

are

a. b.

c. d. None of these.

26. If is the harmonic mean between and then the

value of is

a. 2 b.

c.

d. None of these

27. Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6

and cards are to be placed in envelopes so that each

envelope contains exactly one card and no card is placed

in the enveloped bearing the same number and moreover

the card numbered 1 is always placed in envelope

numbered 2. Then, the number of ways it can be done is

a. 264 b. 265

c. 53 d. 67

28. ABCD is a rectangular field. A vertical lamp post of height

12m stands at the corner A. If the angle of elevation of its

top from B is 60° and from C is 45°, then the area of the

field is

a. 48 2 sq. m b. 48 3 sq. m

c. 48 sq. m

d. 12 2 sq. m

29. The area bounded by the curves and

x-axis in the 1st quadrant is:

a. 9 sq unit

b. 27/4 sq unit

c. 36 sq unit

d. 18 sq unit

30. If the line touches the hyperbola

then the point of contact is:

a. b.

c. d.

0, 2x x= =22 , 2xy y x x= = −

4 1

3 log 2−

3 4

log 2 3+

41

log 2−

3 4

log 2 3−

(sin cos )xdye x x

dx= +

(sin cos )xy e x x c= − +

(cos sin )xy e x x c= − +

sinxy e x c= +

cosxy e x c= +

2 2( ,0)± −a b cos sin 1θ θ+ =x y

a b2a 2b

2 2+a b2 2−a b

2 16y x=2 2( 6) 2,x y− + =

1,1− 2,2−

2,1/ 2− 2, 1/ 2−

, ,a b crr r

,a b a c⋅ = ⋅rr r r

b c=r r

( )a b c⊥ −rr r

b c=r r

( )a b c⊥ −rr r

5 7 2 3 3 6,

3 1 1 36 2 4

x y z x y z− − + + − −= = = =

− −

5 1021, ,

3 3(2,10,4)

( 3,3,6)− (5,7, 2)−

~ ( )∧p q

~ ~∨p q ~ ~∧p q

~ ∧p q ~∧p q

( 1) ( ) 8 3f x f x x+ − = +

( ) 2 ,f x bx cx d= + +

2, 1b c= = 4, 1b c= = −

1, 4b c= − =

H p ,q

H H

p q+

pq

p q+

p q

pq

+

, 2 3y x y x= + =

2 6 2x y+ =2 2

2 4,x y− =

( 2, 6)− ( 5,2 6)−

1 1,

2 6

(4, 6)−

171Mock Test-3

JEE ADVANCE PAPER-I



9 (both inclusive).

1. If z is any complex number satisfying | z 3 2i | 2,− − ≤ then

the minimum value of | 2z 6 5i |− + is

2. Let M be a 3 3× matrix satisfying

0 1 1 1

M 1 2 , M 1 1 ,

0 3 0 1

− = − = −

1 0

and M 1 0 .

1 12

=

Then the sum of the diagonal entries of M is

3. Suppose that all the terms of an arithmetic progression

(A.P.) are natural numbers. If the ratio of the sum of the

first seven terms to the sum of the first eleven terms is

6 : 11 and the seventh term lies in between 130 and 140,

then the common difference of this A.P. is

4. A pack contains n cards numbered from 1 to n. Two

consecutive numbered cards are removed from the pack

and the sum of the numbers on the remaining cards is

1224. If the smaller of the numbers on the removed cards

is k, then

5. Consider a triangle ABC and let a, b and c denote the

lengths of the sides opposite to vertices A, B and C

respectively. Suppose 6, 10a b= = and the area of the

triangle is 15 3. If ∠ACB is obtuse and if r denotes the

radius of the in circle of the triangle, then 2r is equal to

6. Let m and n be two positive integers greater than 1. If

then the value of is

7. Let S be the focus of the parabola y2 = 8x and let PQ be

the common chord of the circle x2 + y

2 – 2x –4y = 0 and

the given parabola. The area of the triangle PQS is.

8. Let be an integer. Take n distinct points on a circle

and join each pair of points by a line segment. Colour the

line segment joining every pair of adjacent points by blue

and the rest by red. If the number of red and blue line

segments are equal, then the value of n is



9. Let A and B be two sets containing four and two elements

respectively. Then the number of subsets of the set

each having at least three elements is:

a. 219 b. 256

c. 275 d. 510

10. The real number for which the equation,

has two distinct real roots in [0, 1]

a. lies between 1 and 2 b. lies between 2 and 3

c. lies between –1 and 0 d. does not exist

11. If is a complex number of unit modulus and argument

then arg equals

a. b.

c. d.

12. If and A adj A = A then is equal

to:

a. –1 b. 5

c. 4 d. 13

13. Three positive numbers form an increasing G.P. If the

middle term in this G.P. is doubled, the new numbers are

in A.P. Then the common ratio of the G.P. is :

a.

b.

c. d.

14. If the number of terms in the expansion of

is 28, then the sum of coefficients

of all the terms in this expansion, is:

a. 64 b. 2187

c. 243 d. 729

k 20 _____− =

cos( )

0lim

2

n

ma

e e eα

α→

− = −

m

n

2n ≥

,×A B

k

32 + 3 0x x k+ =

z

,θ1

1

z

z

+ +

θ−2

πθ−

θ π θ−

5

3 2

a bA

− =

,TA 5a b+

2 3+ 3 2+

2 3− 2 3+

2

2 41 , 0,

n

xx x

− + ≠

Q

P

S

(2,0)

Mathematics172

15. Let be the number of all possible triangles formed by

joining vertices of a n-sides regular polygon. If

then the value of n is

a. 7 b. 5

c. 10 d. 8

16. A computer producing factory has only two plants and

Plant produces 20% and plant produces 80%

of the total computers produced. 7% of computers

produced in the factory turn out to be defective. It is

known that P(computer turns out to be defective given

that it is produced in plant (computer turns out

to be defective given that it is produced in Plant

where P(E) denotes the probability of an event E. A

computer produced in the factory is randomly selected and

it does not turn out to be defective. Then the probability

that it is produced in plant is

a. b.

c. d.

17. A bird is sitting on the top of a vertical pole 20 m high and

is elevation from a point O on the ground is . It flies

off horizontally straight away from the point O. After one

second, the elevation of the bird from O is reduced to

Then the speed (in m/s) of the bird is:

a.

b.

c.

d.

18. Let be differentiable on the interval such that

and for each

Then is

a.

b.

c.

d.



Column II.

19. Match the statements given in Column I with the values

given in Column II.

Column I Column II

(A) If ˆ ˆ ˆ ˆa j 3k j 3k= + = − +r

and ˆc 2 3k=r

form a

triangle, then the internal

angle of the triangle

between ar

and br

is

1. 6

π

(B) If

b

2 2

a

(f (x) 3x)dx a b ,− = −∫

then the value of f6

π

is

2. 2

3

π

(C) The value of 2

ln 3

π

5 / 6

7 / 6

sec( x) dxπ∫ is

3. 3

π

(D) The maximum value

1Arg

1 z

−

for

| z | 1, z 1= ≠ f is given by

4. π

5.

2

π

20. Consider the lines given by1 : 3 5 0,L x y+ − =

2 : 3 1 0,L x ky− − = 3 : 5 2L x y+ − 12 0= Match the

Statements /Expressions in Column I with the Statements/

Expressions in Column II

Column I Column II

(A) 1 2 3, ,L L L are

concurrent, if

1. 9k = −

(B) One of 1 2 3, ,L L L is

parallel to at least of

the other two, if

2. 6

5k = −

(C) 1 2 3, ,L L L form a

triangle, if 3.

5

6k =

(D) 1 2 3, ,L L L do not form

a triangle, if

4. 5k =

nT

1 1n nT T+ − =

1T

2.T 1T 2T

1)T 10 P=

2 ),T

2T

36

73

47

79

78

93

75

83

45°

30 .°

40( 2 1)−

40( 3 2)−

20 2

20( 3 1)−

( )f x (0, )∞

(1) 1,f =2 2( ) ( )

lim 1→

−=

−t x

t f x x f t

t x0.>x

( )f x

21 2

3 3+

x

x

21 4

3 3− +

x

x

2

1 2− +

x x

1

x

173Mock Test-3




9 (both inclusive).

1. For a point P in the plane, let and be the

distances of the point P from the lines and

respectively. The area of the region R

consisting of all points P lying in the first quadrant of the

plane and satisfying is _______

2. The slope of the tangent to the curve

at the point (1, 3) is

3. Let (x, y, z) be points with integer coordinates satisfying

the system of homogeneous equations: 3 0− − =x y z ,

3 0− + =x z 3 2 0.− + + =x y z

Then the number of such points for which

2 2 2 100+ + ≤x y z is

4. Let 1 sin

f ( ) sin tan ,cos2

θθ

θ−

=

where .4 4

π πθ− < <

Then the value of d

(f ( ))d(tan )

θθ

is

5. Two parallel chords of a circle of radius 2 are at a distance

3 1+ apart. If the chords subtend at the centre, angles of

k

πand

2,

k

πwhere 0,k > then the value of [k] is [Note: [k]

denotes the largest integer less than or equal to k].

6. Let a, b, c be positive integers such that is an integer.

If a, b, c are in geometric progression and the arithmetic

mean of a, b, c is b + 2, then the value of is

7. Let i / 3e ,πω = and a, b, c, x, y, z be non-zero complex

numbers such that

a b c x+ + =

2a b c yω ω+ + =

2a b c z.ω ω+ + =

Then the value of 2 2 2

2 2 2

| x | | y | | z |

| a | | b | | c |

+ ++ +

is

8. If the normals of the parabola drawn at the end

points of its latus rectum are tangents to the circle

then the value of is



9. Let then log is equal to:

a. 2 b. 1 c. d.

10. Let be a non-constant twice differentiable function

defined on such that and

Then

a. vanishes at least twice on [0, 1]

b.

c.

d.

11. The following integral is equal to

a.

b.

c.

d.

12. is equal to

a. b.

c. d.

1 ( )d P 2 ( )d P

0x y− =

0x y+ =

1 22 ( ) ( ) 4,d P d P≤ + ≤

5 2 2 2( ) (1 )y x x x− = +

b

a

214

1

a a

a

+ −

+

24=y x

2 2 2( 3) ( 2) ,− + + =x y r2r

1

2 2

0lim (1 tan ) x

xp x

→ += + p

1

2

1

4

( )f x

( , )−∞ ∞ ( ) (1 )f x f x= −

10.

4f ′ =

( )′′f x

10

2f ′ =

1/ 2

1/ 2

1sin 0

2f x x dx

−

+ =

∫

( )1/ 2 1

sin sin

0 1/ 2

(1 1)t tf t e dt f e dtπ π= −∫ ∫

/ 2

17

/ 4

(2cos )

π

π∫ ec x dx

log(1 2 )

16

0

2( )

+−+∫ u ue e du

log(1 2)

17

0

( )

+−+∫ u ue e du

log(1 2 )

17

0

( )

+−−∫ u ue e du

log(1 2 )

16

0

2( )

+−−∫ u ue e du

2

3 4 2

1

2 2 1

xdx

x x x

−

− +∫

4 2

2

2 2 1x xc

x

− ++

4 2

3

2 2 1x xc

x

− ++

4 22 2 1x x

cx

− ++

4 2

2

2 2 1

2

x xc

x

− ++

Mathematics174

13. If y (x) satisfies the differential equation

' tan 2 secy y x x x− = and y(0) = 0, then.

a. 2

4 8 2yπ π =

b. 2

'4 18

yπ π =

c. 2

3 9yπ π =

d. 2

4 2'

3 3 3 3y

π π π = +

14. Given an isosceles triangle, whose one angle is and

radius of its incircle Then the area of the triangle in

sq. units is

a.

b.

c.

d.

15. The circle passing through the point (–1, 0) and touching

the y-axis at (0, 2) also passes through the point

a.

b.

c.

d. (–4, 0)

16 The normal at a point P on the ellipse 2 2x 4y 16+ = meets

the x-axis at Q. If M is the mid-point of the line segment

PQ, then the locus of M intersects the latus rectums of the

given ellipse at the points

a. 3 5 2

,2 7

± ±

b. 3 5 19

,2 4

± ±

c. 1

2 3,7

± ±

d. 4 3

2 3,7

± ±





Tangents are drawn from the point P(3, 4) to the ellipse

touching the ellipse at points A and B.

17. The coordinates of A and B are

a. (3, 0) and (0, 2)

b. and

c. and (0, 2)

d. (3, 0) and

18. The equation of the locus of the point whose distances

from the point P and the line AB are equal, is

a.

b.

c.

d.


Let where

and

19. Area of S =

a. b.

c. d.

20.

a.

b.

c.

d.

120°

3.=

7 12 3+

12 7 3−

12 7 3+

4π

3, 0

2

−

5, 2

2

−

3 5,

2 2

−

2 2

19 4

x y+ =

8 2 161,

5 15

−

9 8,

5 5

−

8 2 161,

5 15

−

9 8,

5 5

−

2 29 6 –54 –62 241 0x y xy x y+ − + =2 2+ 9 6 –54 62 –241 0x y xy x y+ + =

2 29 9 –6 –54 –62 – 241 0x y xy x y+ =2 2+ –2 27 31 –120 0x y xy x y+ + =

1 2 3 ,= ∩ ∩S S S S 1 :| | 4,= ∈ <S z C z

2S =1 3

: Im 01 3

z iz C

i

− + ∈ >

−

3 : Re 0.= ∈ >S z C Z

10

3

π 20

3

π

16

3

π 32

3

π

min |1 3 |∈

− − =z S

i z

2 3

2

−

2 3

2

+

3 3

2

−

3 3

2

+

175Mock Test-3

ANSWER & SOLUTIONS

JEE-Main

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

b c d a a b d a b b

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

b a c a d c d c b c

21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

a c a a b a c a a d

1. (b)

Since

Hence

2. (c)

⇒

3. (d)

or

For

For

4. (a) Let The matrix of cofactors of the elements of A viz.

transpose of the matrix of cofactors of elements

of

.

5. (a) Series is a G.P. with and

6. (b) L.H.S.

7. (d) Let and let

nth

term of given series

or

Now, sum

8. (a) Words starting from are

Words starting from I are 5 ! 120

Words starting from KA are 4 ! 24

Words starting from KI are 4 ! 24

Words starting from KN are 4 ! 24

Words starting from KRA are 3 ! 6

Words starting from KRIA are 2 ! 2

Words starting from KRIN are 2 ! 2

Words starting from KRISA are 1 ! 1

Words starting from KRISNA are 1 ! 1

Hence rank of the word KRISNA is 324.

9. (b) The number of ways of placing 3 black balls without

any restriction is Since, we have total 10 places of

putting 10 balls in a row. Now the number of ways in

which no two black balls put together is equal to the

number of ways of choosing 3 places marked out of

eight places.

This can be done in ways.

Required probability

10. (b)

11. (b) Expression

where

1 ( ) 0

( ( )) 0 ( ) 0

1 ( ) 0

g x

f g x g x

g x

− <

= = >

( ) 1 0g x ≥ >

( ( )) 1g g x =

6 ,x x= + 20 6 , 0x x x x> ⇒ = + >

26 0, 0x x x− − = >

⇒ 3, 0.x x= >

1z z−+ 1= ⇒2

1 0z z− + =

⇒ z ω= −2ω−

,z ω= −100 100 100 100( ) ( )z z ω ω− −+ = − + −

211ω ω ω

ω= + = + = −

2,z ω= − 100 100 2 100 2 100( ) ( )z z ω ω− −+ = − + −

200

200

1ω

ω= + 2 2

2

1ω ω ω

ω= + = +

1.= −

11 12

21 22

2 ( 4) 2 4

( 3) 2 3 2

c c

c c

− − = = − −

∴ adjA =

2 3

4 2A

=

∴ ( ) | |A adj A A I=

90.9

10a = =

10.1

10r = =

∴100 100

100 100

11

1 9 110 1 .11 10 10

110

rS a

r

− −

= = = − − −

0 1 2 3[ ...( 1) . ]n

na C C C C C= − + − + −

1

1 2 3[ 2 3 .... ( 1) . ]n

nC C C nC−+ − + − + − .0 0 0a= + =

2 6 12 20

1! 2! 3! 4!S = + + + +K

1 2 6 12 20n

S T= + + + + +K

1 12 6 12n n

S T T−= + + + +K

0 2 4 6 8 upto termsn

n T= + + + + −K

2 4 6 8 .......upto termsnT n= + + + +

⇒nT [2 2 ( 1) 2]

2

nn= × + − (2 1) ( 1)n n n n= + − = +

∴

( 1) ( 1)or

! ( 1)!n n

n n n nT T

n n n

+ += =

−

1 2

( 2)! ( 1)!nT

n n= +

− −

1 1

1 12 2 3 .

( 2)! ( 1)!n n

e e en n

∞ ∞

= =

= + = + =− −

∑ ∑

A 5 ! 120=

=

=

=

=

=

=

=

=

=

10

3.C

' '−

W W W W W W W− − − − − − − −8

3C

∴8

3

10

3

8 7 6 7

10 9 8 15

C

C

× ×= = =

× ×

2

1 cos 1 cos 1 cos 1 cos

1 cos 1 cos 1 cos

α α α α

α α α

− + − + ++ =

+ − −

2

sinα=±

2 3, since .

sin 2

ππ α

α

= < < −

Q1cos (cos sin )iθ θ− +

1sin sin log( sin 1 sin ),iθ θ θ−= − + +6

πθ =

Mathematics176

Real part of

12. (a) 64 cot dθ =

Also (100 64) tan dθ− = or 2(64)(36) d= ,

∴ 8 6 48 .d m= × =

13. (c) and

Hence is continuous at

14. (a) y x yx e −=

⇒ logy x x y= −

⇒ 1 log

xy

x=

+

⇒ 2 2log (1 log ) log [log ] .

dyx x x ex

dx

− −= + =

15. (d) Let

and

⇒

Now then

Now

Hence is the point of maxima

16. (c)

Put

then it reduces to

17. (d) Required area =

18. (c) Given equation

⇒

On integrating, we get

19. (b) 2 2 2 2

2 2 2 2 2 2 2 2

cos 0 cos

cos sin cos sin

b a b ab b a b ab

b a b a

θ θ

θ θ θ θ

− + − − − − + +

2 2 2 2 2 2

2 2 2 2

[ ( )cos ]

( cos sin )

b a b a b

b a

θθ θ

− − −=

+

2 2 2 2 2 2

2 2 2 2

[ cos cos ]

cos sin

b a a b

b a

θ θθ θ

− +=

+2 2 2 2 2

2

2 2 2 2

[ sin cos ]

cos sin

b a bb

b a

θ θθ θ+

= =+

20. (c)

Let AB be the chord of length O be centre of the circle

and let OC be the perpendicular from O on AB. Then

In

Area of the circle

21. (a)

Here, the focal chord of is tangent to circle

focus of parabola as ie,

Now, tangents are drawn from to

Since, PA is tangent to circle.

slope of tangent or

Slope of focal chord as tangent to circle

22. (c)

Either or

23. (a) Given lines are,

, (say)

∴ 1 13 1cos sin .

2 2 2 4

i π− − + = =

(0 ) (0 ) 2f f+ = − = (0) 2f =

( )f x 0.x =

20 20x y y x+ = ⇒ = −3 2 3 2. .x y z z x y= ⇒ =

3 2(20 )z x x= −3 5 4400 40z x x x= + −

2 4 31200 5 160dz

x x xdx

= + −

0,dz

dx= 12, 20x =

23 2

22400 16 480 ;

d zx x x

dx= + −

2

2

12x

d zive

dx=

= −

12x =∴ 12, 8.x y= =

4 21tan .secx x dx

x∫

2sectan ,

2

xx t dx dt

x= ⇒ =

4 5 52 22 (tan ) tan .

5 5t dt x c x c= + = +∫

22

0[2 (2 )]x x x dx− −∫

23

2

0

2

log 2 3

x xx

= − +

4 8 14

log 2 3 log 2= − + −

3 4.

log 2 3= −

(sin cos )xdye x x

dx= +

(sin cos )x

dy e x x dx= +

sin .xy e x c= +

2 ,

2 1

2 2AC BC= = =

1, cosec 45 . 2 1

2OBC OB BC∆ = ° = =

∴ 2( )OBπ π= =

2 16y x=2 2( 6) 2.x y− + =

⇒ ( ,0)a (4,0)

(4,0) 2 2( 6) 2.x y− + =

∴ tanθ =2

1,2

AC

AP= = = 1

BC

BP= −

∴ 1.= ±

0 ( ) 0a b a c a b a c a b c⋅ = ⋅ ⇒ ⋅ − ⋅ = ⇒ ⋅ − =r r rr r r r r r r r

⇒ 0 or 0b c a b c− = = ⇒ =r rr r r

( ).a b c⊥ −rr r

1

5 7 2

3 1 1

x y zr

− − += = =

−

y

'y

'x x

A

B

θθ

P

(4,0) 2

C(6,0) 2

Tangent as

focal chord

45°

O

A B C

AB = 2

177Mock Test-3

and , (say)

,

and

On solving, we get

Trick: Check through options.

24. (a) .

25. (b) From the given identity

26. (a) As given

.

27. (c) Plan: A square matrix M is invertible if f det(M) or

Let,

(a) Given that (let)

M is non-invertible.

(b) Given that

(let)

Again

M is non-invertible.

(c) As given

( a and c are non-zero)

M is invertible.

(s)

ac is not equal to square of an integer.

M is invertible.

28. (a)

Let AE is a vertical lamp-post. Given, AE = 12m

tan 45AE

AC° =

AC AE m= =12m

tan 60AE

AB° =

4 33

AEAB = =

2 2 144 48 96 4 6BC AC AB= − = − = =

Area = 4 3 4 6 48 2 . .AB BC sq cm= × = × = sq.cm.

29. (a)

To find the area between the curves,

and x-axis in the Ist quadrant (we can plot the above

condition as);

Area of shaded portion OABO

sq unit

30. (d) The equation of tangent at is

which is same as

∴

and

JEE Advance Paper-I

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

5 9 9 5 3 2 4 5 a d

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

c b d d b c d a c a

1. (5) 5

| 2z 6 5i | 2 z 32

− + = − −

5

2 3 3 i2

≥ − −

(corresponding Pt A) 5

2 52

= =

2

3 3 6

36 2 4

x y zr

+ − −= = =

−

∴ 1 23 5 36 3x r r= + = − −

1 27 3 2y r r= − + = +

1 22 4 6z r r= − = +

5 1021, ,

3 3x y z= = =

qpqp ~~)(~ ∨≡∧

2 2( 1) ( 1) ( ) 8 3b x c x d bx cx d x+ + + + − + + = +

⇒ 2 8 3bx b c x+ + = +⇒ 4, 1b c= = −

2 pqH

p q=

+

∴2 2 2( )

2H H q p p q

p q p q p q p q

++ = + = =

+ + +

| | 0.M ≠

a bM

b c

=

a ba b c

b cα

= ⇒ = = =

⇒ | | 0M Mα αα α

= ⇒ =

⇒

[ ] [ ]b c a b=

⇒ a b c α= = =

| | 0M =

⇒

0| | 0

0

aM M ac

c

= ⇒ = ≠

Q

⇒

2| | 0a b

M M ac bb c

= ⇒ = − ≠

Q

∴

, 2 3y x y x= + =

9 9

0 3

3

2

xx dx dx

− = − ∫ ∫

9 93/ 2 2

0 3

13

3/ 2 2 2

x xx

= − −

2 1 81 927 27 9

3 2 2 2

= ⋅ − − − −

118 (18) 9

2= − =

1 1( , )x y 1 12 4,xx yy− =

2 6 2x y+ =

1 12 4

2 26

x y= − =

⇒ 1 4x =1 6y = −

O

'y

y

'x x

B

A(3,0) (9,0)

y x=

3

2

xy

−=

600

E

A B

C

450

D

Mathematics178

2. (9) Let

a b c

M d e f

g h i

=

0 1

M 1 2 b 1, e 2, h 3

0 3

− = ⇒ = − = =

1 1

M 1 1 a 0, d 3, g 2

0 1

− = ⇒ = = = −

1 0

M 1 1 g h i 12 i 7

1 12

= ⇒ + + = ⇒ =

∴ Sum of diagonal elements = 9.

3. (9) Let seventh term be ‘a’ and common difference be‘d’

Given

Hence,

4. (5) Clearly,

and

5. (3)1

2∆ = ab sin C

⇒ 2 2 15 3 3

sin 1206 10 2

C Cab

∆ ×= = = ⇒ = °

×

⇒ 2 2 2 cosc a b ab C= + −

2 26 10 2 6 10 cos120 14= + − × × × ° =

∴ rs

∆= ⇒

2 225 33.

6 10 14

2

r×

= =+ +

6. (2)

if and only if

7. (4) Solving 2 8y x= and

2 2 2 4 0x y x y+ − − =

Simultaneously, we get (2, 4) and (0, 0)

Focus is (2, 0)

∴ Are 1

2 4 4sq.2

= × × = units.

8. (5) Number of red lines

Number of blue lines = n

Hence,

⇒

⇒

9. (a) Set A has 4 elements

Set B has 2 elements

Number of elements in set

Total number of subsets of

Number of subsets having 0 elements

Number of subsets having 1 element each

∴ Number of subsets having 2 elements each

Number of subsets having at least 3 elements

10. (d)

⇒

Not possible. As condition for two distinct real root is

(where are roots of

11. (c) Let

Now

arg (put

12. (b)

7

11

6

11=

S

S

⇒ 15=a d

130 15 140< <d ⇒ 9=d

1 2 3 2 1224 3 4n n+ + + + − ≤ ≤ + +K K

⇒( 2) ( 1) ( 2)

1224 (3 )2 2

n n nn

− − −≤ ≤ +

⇒ 2 3 2446 0n n− − ≤ 2 2454 0n n+ − ≥

⇒ 49 51n< <

⇒ 50n =

∴ ( 1)(2 1) 1224

2

n nk

+− + =

⇒ 25 20 5k k= ⇒ − =

( )cos

0lim

2

α

α α→

−= −

n

m

e e e

( )( )

(cos( ) 1)

20

(cos 1)lim

cos( ) 1

α

α

α

α α α

−

→

−

−

n n

n m n

e e2

2α = −n e

2 0− =n m

2= −nC n

2 − =n C n n

2 2=n C n

( 1)2

2

−=

n nn

1 4− =n

⇒ 5.=n

∴ ( ) 4 2 8× = × =A B

∴ 8( ) 2 256× = =A B

08 1= =C

18 8= =C

2

8! 8 78 28

2!6! 2

×= = = =C

256 1 8 28 256 37 219= − − − = − =3( ) 2 3f x x x k= + +

2( ) 6 3f x x′ = +

( ) 0f x′ = ⇒ 2 1

2x = −

( ) ( ) 0f fα β =

,α β ( ) 0)f x′ =

z ω=2

2

1 1

1 1

z

z

ω ωω

ω ω+ + −

= = =+ + −

∴ 1arg

1

z

zω θ

+= =

+cos sin )z iθ θ= +

| | TA I AA=

⇒1 0 5 5 3

(10 3 )0 1 3 2 2

a b aa b

b

− + = −

⇒ 2 225 10 3a b a b+ = + &15 2 0 &10 3 13a b a b− = + =

⇒3.15

10 132

+ =a

a ⇒ 65 2 13= ×a

179Mock Test-3

13. (d) Let the numbers be a, ar, ar2 is G.P.

Given a, 2ar, ar2 are in A.P. the

which gives as the G.P. is an increasing G.P.

14. (d) Theoretically the number of terms are 2N + 1 (i.e. odd)

But As the number of terms being odd hence considering

that number clubbing of terms is done hence the solutions

follows:

Number of terms =

Sum of coefficient =

Put

15. (b)

⇒ On solving n = 5

16. (c) Let (computer turns out to be defective given

that it is produced in plate

P (produced in not defective)

17. (d)

Here,

Hence distance covered in one second by the bird is

Thus speed of bird

–1) m/s

18. (a)

Also

Hence

19. (A) a b 1 3 2− = + =rr

| a | b, | b | 2= =rr

2 1

cos2 2 2

θ = =×

2

,3 3

π πθ = but its

2

3

πas its opposite to side of maximum

length.

(B)

b

2 2

a

(f (x) 3x) dx a b− = −∫

b 2 22 2 2 2

a

3 a bf (x)dx (b a ) a b

2 2

− += − + − =∫

⇒ f (x) x.=

(C)

5 / 62

7 / 6ln (sec x tan x)

ln 3

π πππ

+

5 5 7 7

ln sec tan ln sec tan .ln 3 6 6 6 6

π π π π ππ

= + − + =

(D) Let 1 1

u z 11 z u

= ⇒ = −−

1

| z | 1 1 12

= ⇒ − =

⇒

| u 1| | u |− =

∴ locus of u is perpendicular bisector of line segment

joining 0 and 1

⇒ maximum arg u approaches 2

πbut will not attain.

⇒2

5=a ⇒ 5 2=a

⇒ 2 6=a ⇒ 3=b

∴ 5 5+ =a b

2

2 ( 0)2

a arar a

+= ≠

2 3,r = +

2

2 28+ =n C

∴ 6=n

63 3 729= =n

1=x

1

3 3 10n nC C+ − =

x P=

2 ),T

⇒7 1 4

(10 )100 5 5

x x= +

⇒ 7 200 80x x= +

⇒7

280x =

2 /T

( )

( )

P A B

P B

∩=

⇒

4 4 273

5(1 ) 5 280

1 4 1 280 70 4 273(1 10 ) (1 )

5 5 5 280 5 280

x

x x

− =

− − + − +

⇒4 273 2 273 546 78

210 4 273 105 2 273 651 93

× ×= = =

+ × + ×

20AP QB m= =

45 ,POA∠ = ° 30QOB∠ = °

⇒ 20;OA= 20 3OB =

⇒ 20( 3 1)OB OA− = −

20( 3 1)AB = −

20( 3 1) /= −

2 2( ) ( )lim 1→

−=

−t x

t f x x f t

t x

⇒ 2 ( ) 2 ( ) 1 0′ − + =x f x x f x

⇒ 2 1( )

3= +f x cx

x

(1) 1=f

⇒ 2

3=c

22 1( ) .

3 3f x x

x= +

Q

B

P

A O

Mathematics180

20. 3 5 0x y+ − = and 5 2 12 0x y+ − = intersect at (2, 1)

Hence 6 1 0k− − = 5k =

for 1 2,L L to be parallel

1 39

3k

k= ⇒ = −−

for 2 3,L L to be parallel

3 6.

5 2 5

kk

− −= ⇒ =

for 6

5, 9,5

k−

≠ − they will form triangle

for 6

5, 9,5

k k−

= = − they will not form triangle


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

6 8 7 1 5 4 3 2 3 All

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

a d a,d c d a d a b c

1. (6)

For

Area of region

sq. units

2. (8)

Now put

and

⇒

⇒ ⇒

3. (7) 3 0− − =x y z 3 0− + =x z

⇒ 0=y and 3=z x

⇒ 2 2 2 2 2 2 2 29 10 100+ + = + = + = ≤x y z x z x x x

⇒ 2

10≤x ⇒ 0, 1, 2, 3= ± ± ±x

There are such seven points.

4. (1) 1 sin

f ( ) sin tan2cos

θθ

θ−

=

1

2

sinsin sin

sin cos 2

θ

θ θ−

= +

2 2 2

sin sin sintan

| cos | cossin (cos sin )

θ θ θθ

θ θθ θ θ= = = =

+ −

⇒ d d(tan )

f ( ) 1d(tan ) d(tan )

θθ

θ θ= =

5 (2) 2cos 2cos 3 12k k

π π+ = +

⇒ 3 1

cos cos2 2k k

π π ++ =

Let 3 1

0,cos cos2 2k

π θθ

+= + =

⇒ 2 3 1

2cos 1 cos2 2 2

θ θ +− + =

⇒ cos2

tθ= ⇒ 2 3 3

2 02

t t+

+ − =

⇒ 1 1 4(3 3) 1 (2 3 1) 2 2 3 3

,4 4 4 2

t− ± + + − ± + − −

= = =

Q 3

[ 1,1],cos2 2

tθ

∈ − =

⇒ 2 6

θ π= ⇒ 3.k =

6. (4)

⇒

only

7. (3) The expression may not attain integral value for all a, b, c

If we consider a = b = c, then x = 3a

2y a(1 ) a(1 i 3)ω ω= + + = +

2z a(1 ) a(1 i 3)ω ω= + + = +

∴ 2 2 2 2 2 2 2| x | | y | | z | 9 | a | 4 | a | 4 | a | 17 | a |+ + = + + =

2=x 2 2=x

( , )α βP

=y x

1 22 ( ) ( ) 4≤ + ≤d p d p

( , ),α β α β>P

⇒ 2 2 2 4 2α≤ ≤

2 2 2α≤ ≤

⇒ 2 2((2 2) ( 2) )= −

8 2 6 .= − =

5 42( ) 5 − −

dyy x x

dx

2 2 21(1 ) ( )(2(1 )(2 ))= + + +x x x x

1, 3= =x y

.=dy

mdx

2(3 1)( 5) 1(4) (1)(4)(2)− − = +m

125

4− =m

5 3 8= + =m 8.= =dy

mdx

(integer)= =b c

a b

2 =b ac

⇒2

=b

ca

23

+ += +

a b cb

3 6+ + = +a b c b

⇒ 2 6− + =a b c

2

2 6− + =b

a ba

⇒2

2

2 61− + =

b b

a aa

26

1 − =

b

a a6=a

181Mock Test-3

8. (2) Equation of normals are and

Distance from on both normals is ‘r’

9 (3) then

⇒

10. (a,b,c,d)

Put

⇒

Hence is an even function or

an odd function.

Also, and for

We have

Also,

(obtained by putting,

Since Also

atleast twice in [0, 1] (Rolle’s Theorem)

11. (a)

Let

and

12. (d)

Let

⇒

13. (a,d) ' tan 2 secy y x x x− =

tan log cosI.F. cos

x dx xe e x∫= = =

∴ cos 2 sec .cosy x x x x dx= ∫

⇒ 2cosy x x c= +

⇒ 2cos ( (0) 0)y x x y⋅ = =Q

⇒ 2 secy x x=

∴ 2 2

4 2and y'

4 3 38 2 3 3yπ π π π π = = +

14. (c) . . .(i)

Also

and and

. . .(ii)

From equation (i) and (ii), we get

15. (d) Circle touching y-axis at (0, 2) is

passes through (–1, 0)

Put

Circle passes through (–4, 0)

3x y+ = 3.x y− =

⇒ (3, 2)−

⇒| 3 2 3 |

2r

− −=

⇒2 2.r =

1

2lim (1 tan 2 1)+→∞

= + − x

xP x log =p

212(1 tan 1)2 2

00

(tan ) 1limlim2( ) 2

+ −++ →→= = =

xx

xx

x

xP e e e

1

21

log P loge2

= =

( ) (1 )f x f x= −

1/ 2x x= +

f1 1

2 2x f x

+ = −

( 1/ 2)f x + ( 1/ 2)sinf x x+

'( ) '(1 )f x f x= − − 1/ 2,x =

'(1/ 2) 0.f =1 0

sin sin

1/ 2 1/ 2

(1 ) ( )π π− = −∫ ∫t tf t e dt f y e dy

1 ).t y− =

(1/ 4) 0, (3/ 4) 0.′ ′= =f f (1/ 2) 0′ =f

⇒ ( ) 0f x′ =

217

4

(2cosec )x dx

π

π∫

2cosec ,4

u ue e x xπ−+ = =

⇒ ln(1 2),2

u xπ

= + =

⇒ 0u =

⇒ cosec cot ux x e+ = cosec cot ux x e−− =

⇒ cot2

u ue ex

−−( ) 2cosec cotu ue e dx x xdx−− = −

⇒ 17 ( )( )

2cosec cot

u uu u e e

e e dux x

−− −

− +∫0

16

ln(1 2 )

2 ( )u ue e du−

+

= − +∫ln(1 2 )

16

0

2( )u u

e e du

+−= +∫

3 5

2 4

1 1

2 12

dxx x

x x

−

− +∫

2 4

2 12 z

x x− + =

1

4

dz

z∫

⇒1

2z c× +

⇒2 4

1 2 12

2c

x x− + +

23

4b∆ =

sin120 sin30

a b

° °=

⇒ 3a b= 3s∆ =1

( 2 )2

s a b= +

⇒3

( 2 )2

a b∆ = +

(12 7 3)∆ = +

2 2(x 0) (y 2) x 0λ− + − + =

∴ 1 4 0 5λ λ+ − = ⇒ =

∴ 2 2x y 5x 4y 4 0+ + − + =

y 0= ⇒ x 1, 4= − −

∴

1/4 3/4

1/2

Mathematics182

16. (a) Any point on the line can be taken as

(1 3 ), ( 1), (5 2)µ µ µ≡ − − +Q

3 2, 3, 5 4µ µ µ= − − − −PQ

Now 1( 3 2) 4( 3) 3(5 4) 0µ µ µ− − − − + − =

⇒ 3 2 4 12 15 12 0µ µ µ− − − + + − =

8 2µ = ⇒ 1/ 4µ =

17. (d)

Equation is

Let

⇒

⇒

⇒

18. (a) Locus is parabola

Equation of AB Is

19. (b)

Area of region shaded area

20. (c) Distance of (1, –3) from

29 4y mx m= + +

24 3 9 4m m− = +

2 2 12 116 9 24 9 4

24 2m m m m+ − = + ⇒ = =

14 ( 3)

2y x− = −

2 8 3− = −y x

⇒ 2 5 0x y− + =

( , )B α β=

⇒ 1 09 4

α β+ − =

x y

/ 9 / 4 1

1 2 5

α β −= = ⇒

−9

,5

α = −8

5β =

9 8, .

5 5B

≡ −

3 41

9 4

x y+ =

⇒ 1 3 3 03

xy x y+ = ⇒ + − =

22 2 ( 3 3)

( 3) ( 4)10

x yx y

+ −− + − =

2 2 2 210 90 – 60 10 160 – 80 9 9 6 – 6 – 18x x y y x y xy x y+ + + = + + +2 2 2 210 90 – 60 10 160 – 80 9 9 6 – 6 – 18x x y y x y xy x y+ + + = + + +

⇒ 2 2 9 – 6 – 54 – 62 241 0.x y xy x y+ + =

1 2 3∩ ∩ =S S S

2 24 4

4 6

π π× ×= +

2 1 14

4 6π = +

20

3

π=

3 0+ =y x

3 3 1

2

− + ×>

3 3

2

−>

3 0+ =y x

2 2 16+ <x y60°

60°

183Mock Test-4


1. The number of solutions of is:

a. 3 b. 1 c. 2 d. 0

2. The real roots of the equation are

a. – 1, 4 b. 1, 4

c. – 4, 4 d. None of these

3. If is a root of equation then

its real roots are

a. 1, 1 b. – 1, – 1 c. 1, – 1 d. 1, 2

4. Let , then the adjoint of A is

a. b.

c.

d. None of these

5. The value of is

a. b.

c. d.

6. The sum of the coefficients of even power of x in the

expansion of is

a. 256 b. 128 c. 512 d. 64

7. In the expansion of the coefficient of

will be

a. b.

c. d.

8. The total number of seven digit numbers the sum of

whose digits is even is

a. 9000000 b. 4500000


9. If E and F are events with and

then

a. occurrence of E occurrence of F

b. occurrence of F occurrence of E

c. non-occurrence of E non-occurrence of F

d. None of the above implication holds

10.

a. b. c. d.

11. equals

a. b.

c. d.

12. The length of the shadow of a pole inclined at 10° to the

vertical towards the sun is 2.05 metres, when the elevation

of the sun is 38°. The length of the pole is

a. 2.05 sin 38

sin 42

°°

b. 2.05 sin 42

sin 38

°°

c. 2.05 cos 38

cos 42

°°

d. None of these

13. If then

a. b.

c. f is continuous at d. None of these

14. If ( )( )

,( )( )

x a x by

x c x d

− −=

− − then

dy

dx=

a. 1 1 1 1

2

y

x a x b x c x d

+ − − − − − −

b. 1 1 1 1

yx a x b x c x d

+ − − − − − −

c. 1 1 1 1 1

2 x a x b x c x d

+ − − − − − −

d. None of these

15. If

2

2

1( ) ,

1

xf x

x

−=

+ for every real number x, then the

minimum value of f

a. Does not exist because f is unbounded

b. Is not attained even though f is bounded

c. Is equal to 1

d. Is equal to –1

log 4( 1) log 2( 3)x x− = −

2 5 | | 4 0x x+ + =

1 3

2

i+4 3 1 0x x x− + − =

1 0 0

5 2 0

1 6 1

A

= −

2 5 32

0 1 6

0 0 2

− −

1 0 0

5 2 0

1 6 1

− − − −

1 0 0

5 2 0

1 6 1

− − − − −

0.234

232

990

232

9990

232

990

232

9909

2(1 x x+ + 3 5)x+

21 2 3,

x

x x

e

− + 5x

71

120

71

120−

31

40

31

40−

( ) ( )P E P F≤ ( ) 0,P E F∩ >

⇒

⇒

⇒

tan tan4 4

π πθ θ + − − =

2 tan 2θ 2cot 2θ tan 2θ cot 2θ

tanh( )x y+

tanh tanh

1 tanh tanh

x y

x y

+−

tanh tanh

1 tanh tanh

x y

x y

++

tanh tanh

1 tanh tanh

x y

x y

−−

tanh tanh

1 tanh tanh

x y

x y

−+

2 1sin , when 0

( ) ,

0, when 0

x xf x x

x

≠=

=

(0 0) 1f + = (0 0) 1f − =

0x =

Mathematics184

16.

a. b.

c. d.

17. The area bounded by the circle line

and x-axis lying in the first quadrant, is

a. b.

c. d.

18. The general solution of is

a. b.

c. d.

19. The length of perpendicular from the point

upon the straight line

is

a. b. c. d.

20. The circle passing through point of intersection of the

circle 0S = and the line 0P = is

a. 0S Pλ+ =

b. 0S Pλ− = and 0S Pλ + =

c. 0P Sλ− =

d. All of these

21. Axis of a parabola is and vertex and focus are at a

distance and respectively from the origin. Then

equation of the parabola is-

a. b.

c. d.

22. If and be unlike vectors, then

a. b.


23. A line makes the same angle with each of the x and

z-axis. If the angle which it makes with y-axis is such

that then equals

a. b.

c. d. None of these

24. is equal to

a. b. c. d.

25. If and are the roots of the equation

and then

a. –8 b. –16 c. 16 d. 8

26. If are in H.P., then is equal to

a. b.

c. d. None of these

27. If then the value of x in terms of

y is

a. b.

c. d.

28. The number of arrangements of the letters of the word

BANANA in which the two N’s do not appear adjacently,

is

a. 40 b. 60

c. 80 d. 100

29. Differential coefficient of1sin x−

w.r.t 1 2cos 1 x− − is

a. 1 b. 2

1

1 x+


30. The interval in which the function 2 xx e− is non decreasing, is

a. ( , 2]−∞ b. [0, 2]

c. [2, )∞ d. None of these

21

x

x

adx

a=

−∫

11sin

log

xa ca

− + 1sin xa c− +

11cos

log

xa ca

− + 1cos xa c− +

2 24,x y+ = 3x y=

2

π4

π3

π π

2 2dy

xdx

=

2y c

x= +

2y c

x= − 2y cx=

2

3y c

x= −

( cos , sin )α αa a

tan , 0α= + >y x c c

cosαc 2sin αc 2sec αc 2cos αc

y x=

2 2 2

2( ) 8( 2)x y x y− = + − 2( ) 2 ( 2)x y x y+ = + −2( ) 4 ( 2)x y x y− = + − 2( ) 2 ( 2)x y x y+ = − +

ar

br

a b⋅ =rr

| | | |a brr

| | | |a b−rr

,θ

,β2 2sin 3sin ,β θ= 2

cos θ

3

5

2

3

1

5

(~ (~ )) ∧p q

qp ∧~ ∧p q ~∧p q ~ ~∧p q

α β2 6 0x x λ+ + = 3 2 20,α β+ = − λ =

, , ,a b c d ab bc cd+ +

3ad ( )( )a b c d+ +

3ac

2 33 6 10 ....,y x x x= + + +

1/ 31 (1 )y −− − 1/ 31 (1 )y− +1/ 3

1 (1 )y−+ + 1/ 3

1 (1 )y−− +


185Mock Test-4

JEE ADVANCE PAPER-I



9 (both inclusive).

1. For any integer k, let where

The value of the expression is

2. Let m be the smallest positive integer such that the

coefficient of 2x in the expansion of

2 3 49 50(1 ) (1 ) ... (1 ) (1 )x x x mx+ + + + + + + + is

51

3(3 1)n C+

for some positive integer n. Then the value of n is

3. The minimum number of times a fair coin needs to be

tossed, so that the probability of getting at least two heads

is at least 0.96 is

4. For any real number x, let |x| denote the largest integer less

than or equal to x. Let f be a real valued function defined

on the interval [−10, 10] by f (x) = [ ] if [ ] is odd

1 [ ] if [ ] is even

x x x

x x x

-ÏÌ+ -Ó

Then the value of is

5. The value of is ________

6. Let f be a real-valued differentiable function on R (the set

of all real numbers) such that f(1) = 1. If the y-intercept of

the tangent at any point P(x, y) on the curve y = f(x) is

equal to the cube of the abscissa of P, then the value of

f (−3) is equal to

7. The line is tangent to the hyperbola

If this line passes through the point of

intersection of the nearest directrix and the x-axis, then the

eccentricity of the hyperbola is

8. If ,a brr

and cr

are unit vectors satisfying

2 2 2

9,a b b c c a− + − + − =r rr r r r

then 2 5 5a b c+ +rr r

is.



9. If then at is equal to

a. b. c. 1 d.

10. For function f(x)

a. for atleast one x in interval

b.

c. for all x in the interval

d. f’ (x) is strcily decreasing in the interval

11. Let 1/

( )(1 )

=+ n n

xf x

xfor 2≥n and

occurs times

( ) ( )( ).= o oKo1442443

f n

g x f f f x Then 2 ( )−∫ nx g x dx equals

a.

111

(1 )( 1)

−+ +

−n nnx K

n n b.

111

(1 )1

−+ +

−n nnx K

n

c.

111

(1 )( 1)

n nnx Kn n

++ +

+ d.

111

(1 )1

n nnx Kn

++ +

+

12. Let be a function which is continuous on

[0, 2] and is differentiable on (0, 2) with Let

for If for all

then equals

a. b. c. d.

13. The function is the solution of the differential

equation in (–1,1) satisfying

Then is

a. b. c. d.

14. Let (0, 0), (3, 4), (6, 0)O P Q be the vertices of the triangle

OPQ. The point R inside the triangle OPQ is such that the

triangles OPR, PQR, OQR are of equal area. The

coordinates of R are

cos sin ,7 7

k

k ki

π πα

= +

1.i = −

12

11

3

4 1 4 21

| |

| |

k kk

k kk

α α

α α

+=

− −=

Σ −

Σ −

102

10

( )cos10

f x xdxπ

π−∫

1 23 2 5

2

0

4 (1 )

−

∫d

x x dxdx

2 1x y+ =

2 2

2 21.

x y

a b− =

1sec(tan ),y x−= dy

dx1x =

1

2

1

22

1x cos ,x 1,

x= ≥

[1, ),f (x 2) f (x) 2∞ + − <

xlim f '(x) 1

→∞=

[1, ), f (x 2) f (x) 2∞ + − >

[1, )∞

: [0,2]f R→

(0) 1.=f

2

0

( ) ( )

x

F x f t dt= ∫ [0,2].∈x ( ) ( )′ ′=F x f x

(0,2),∈x (2)F

21−e

41−e 1−e

4e

( )=y f x

2 1+ =

−dy xy x x

dx x

4

2

2

1 1

++ =

−

dy xy x x

x

(0) 0.=f

3

2

3

2

( )∫ f x dx

3

3 2

π−

3

3 4

π−

3

6 4

π−

3

6 2

π−

Mathematics186

a. 4

, 33

b. 2

3,3

c. 4

3,3

d. 4 2

,3 3

15. Tangents are drawn to the hyperbola parallel

to the straight lien The points of contact of the

tangents on the hyperbola are.

a. b.

c. d.

16. If the vectors and are the

sides of a triangles then the length of the median

through is

a. b.

c. d.

17. In consider the planes and Let

be a plane, different from and which passes

through the intersection of and If the distance of the

point (0, 1, 0) from is 1 and the distance of a point

from is 2, then which of the following

relations is (are) true?

a.

b.

c.

d.

18. The negation of is equivalent to

a. b.

c. d.



Column II.

19. Match the statements given in Column I with the

interval/union of intervals given in Column II

Column I Column II

(A) The set 2

2izRe : z

1 z

−

is

a complex number,

| z | 1, z 1= ≠ ± is

1. ( , 1) (1, )−∞ − ∪ ∞

(B) The domain of the

function 1f (x) sin −=

x 2

2(x 1)

8(3)

1 3

−

−

− is

2. ( , 0) (0, )−∞ ∪ ∞

(C) If

1 tan 1

f ( ) tan 1 tan ,

1 tan 1

θθ θ θ

θ= −

− −

then the set

f ( ) : 02

πθ θ ≤ <

is

3. [2, )∞

(D) If 3 / 2f (x) x (3x 10),= −

x 0,≥ then f(x) is

increasing in

4. ( , 1] [1, )−∞ − ∪ ∞

5. ( ,0] [2, )−∞ ∪ ∞

20. Let Match the conditions/expressions

in Column I with statements in Column II

Column I Column II

(A) If then

satisfies

1.

(B) If then

satisfies

2.

(C) If then

satisfies

3.

(D) If then

satisfies

4.

2 2

1,9 4

x y− =

2 1.x y− =

9 1,

2 2 2

9 1,

2 2 2

−

3 3, 2 2− 3 3,2 2−

ˆˆ3 4AB i k= + ˆˆ ˆ5 2 4AC i j k= − +

,ABC

A

18 72

33 45

3,R 1 : 0=P y 2 : 1.+ =P x z

3P 1P 2 ,P

1P 2 .P

3P

(α, β, γ) 3P

2α + β + 2γ 2 0+ =

2α – β + 2γ 4 0+ =

2α + β – 2γ – 10 0=

2α – β + 2γ – 8 0=

~ (~ )s r s∨ ∧

~s r∧ ( ~ )s r s∧ ∧

( ~ )s r s∨ ∨ s r∧

2

2

6 5( )

5 6

x xf x

x x

− +=

− +

1 1,x− < <

( )f x

0 ( ) 1f x< <

1 2,x< <

( )f x

( ) 0f x <

3 5,x< <

( )f x

( ) 0f x >

5,x > ( )f x ( ) 1f x <


187Mock Test-4




9 (both inclusive).

1. The least period of the function

is then the value

of must be (where denotes the greatest integer

function)

2. If 1tan

ax by

bx a

− − = + then the value of

1

(2008)x

dy

dx =−

must be

3. The indicated horse power I of an engines is calculated

from the formula where

Assuming that error of 10% may have been made in

measuring P, L, N and d. If the greatest possible in I is

then must be

4. If

then the value of must be

5. A particle moves in a straight line with a velocity given

by (x is the distance travelled). If the time

taken by a particle to traverse a distance of 99 m is


6. If a triangle has its orthocenter at (1, 1) and circum

centre at and if centroid and nine point centre are

and respectively, then the value of

must be

7. If the radius of a circle which passes through the point

and whose centre is the limit of the point of

intersection of the lines and

as is then the value of

must be

8. A, B, C, D are any four points in the space. If

(area of )




9. Let and be two sets containing 2 elements and 4

elements respectively. The number of subsets of

having 3 or more elements is

a. 256 b. 220

c. 219 d. 211

10. The real number for which the equation,

has two distinct real roots in [0, 1]

a. lies between 1 and 2

b. lies between 2 and 3

c. lies between –1 and 0

d. does not exist

11. A complex number z is said to be unimodular if

Suppose and are complex numbers such that

is unimodular and is not unimodular. Then

the point lies on a

a. straight line parallel to x-axis

b. straight line parallel to y-axis

c. circle of radius 2

d. circle of radius

12. If the adjoint of a 3 3× matrix P is

1 4 4

2 1 7 ,

1 1 3

then the

possible value(s) of the determinant of P is (are)

a. –2 b. –1

c. 1 d. 2

13. If the sum of the first ten term of the series

2 2 23 2 1

1 2 35 5 5

+ + +

24 +2

44 ...,

5

+

is then

m is equal to:

a. 102 b. 101

c. 100 d. 99

14. Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6

and cards are to be placed in envelopes so that each

envelope contains exactly one card and no card is placed

in the envelope bearing the same number and moreover

the card numbered 1 is always placed in envelope

numbered 2. Then the number of ways it can be done is

a. 264 b. 265 c. 53 d. 67

[ ] [ ]sin cos tan

12 4 3

x x xπ π π + +

,λ

201λ [ ]⋅

33000

PLANI = 2.

4A d

π=

%λ λ

1 1

2

2 2 2 2sin ( 1) tan

3(4 8 13)

x xdx x

x x

− − + + = + + +

∫2

ln(4 8 13) ,x x cλ+ + + + 4λ−

( 1)dx

xdt

= +

.λ

1020 log eλ

3 3,

2 4

( , )α β ( , )γ δ

6 12 4 8α β γ δ+ + +

(2, 0)

3 5 1x y+ =

2(2 ) 5 1c x c y+ + = 1c → ,25

λ

λ

| |AB CD BC AD CA BD λ× + × + × =uuur uuur uuur uuur uuur uuur

ABC∆

125λ

A B

A B×

k

32 + 3 0x x k+ =

| | 1.=z

1z 2z

1 2

1 2

2

2

−−

z z

z z2z

1z

2

16,

5m

Mathematics188

15. Let , 0,2θ ϕ π∈ be such that 2cos 1 sinθ ϕ−

2sin tan cot cos 12 2

θ θθ ϕ = + −

tan 2 0π θ− > and

31 sin

2θ− < < − .

Then ϕ cannot satisfy:

a. 02

πϕ< <

b. 4

2 3

π πϕ< <

c. 4 3

3 2

π πϕ< <

d. 3

22

πϕ π< <

16. If then 1/ 2

2 1 1 21 cos(cot ) sin(cot ) 1x x x x− − + + −

is equal to

a. b.

c. d.






Let denote the number of all n-digit positive integers formed

by the digits 0, 1 or

both such that no consecutive digits in them are 0. Let = the

number of such n-digit

integers ending with digit 1 and = the number of such

n–digit integers ending with digit 0.

17. Which of the following is correct?

a. b.

c. d.

18. The value of is

a. 7 b. 8

c. 9 d. 11


Box 1 contains three cards bearing numbers 1, 2, 3; box 2

contains five cards bearing numbers 1, 2, 3, 4, 5: and box 3

contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card

is drawn from each of the boxes/ Let be the number on the

card drawn from the box,

19. The probability that is odd, is

a. b.

c. d.

20. The probability that are in an arithmetic

progression, is

a. b.

c. d.

0 1,x< <

21

x

x+x

21x x+ 21 x+

na

nb

nc

17 16 15a a a= + 17 16 15c c c≠ +

17 16 16b b c≠ + 17 17 16a c b= +

6b

ix

thi 1, 2,3.=i

1 2 3,+x x x

29

105

53

105

57

105

1

2

1 2 3, ,x x x

9

105

10

105

11

105

7

105


189Mock Test-4

ANSWER & SOLUTIONS

JEE-Main

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

b d c d a c b b d a

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

b a c a d a c b a d

21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

a b a b b a d a a b

1. (b) Given

But at given log is not defined.

2. (d)

which is not possible. Hence, the given

equation has no real root.

3. (c)

or

So its real roots are 1 and

4. (d)

.

5. (a)

=

6. (c)

Therefore the required sum of coefficients

Note: Sum of all the binomial coefficients in

the 2nd

bracket in which all the powers of x are even.

7. (b)

∴ The coefficient of

8. (b) Suppose represents a seven digit number.

Then takes the value and all

take values 0, 1, 2, 3, , 9. If we keep fixed,

then the sum is either even or odd. Since

takes 10 values 0, 1, 2, ,9, five of the numbers so

formed will be even and 5 odd.

Hence the required number of numbers

9. (d) It is given that

and . . .(ii)

(a) Occurrence of E occurrence of F [from Eq.(i)]

(b) Occurrence of F occurrence of E [from Eq.(ii)]

(c) Non-occurrence of E non-occurrence of F [from

Eq.(i)]

10. (a)

.

11. (b) It is understandable.

12. (a)

sin 38 sin( )

2.05

o SPO

l=

= sin(180 38 90 10 )

2.05

o o o o− − −⇒

2.05sin 38

sin 42

o

ol =

log( 1) log( 3)

2 log 2 log 2

x x− −=

⇒ 2( 1) ( 3)x x− = −

⇒ 2,5x =

2,x =

25 | | 4 0x x+ + =

⇒ 2| | 5 | | 4 0x x+ + =

⇒ | | 1, 4,x = − −

4 3 1 0x x x− + − =

⇒3 ( 1) 1( 1) 0x x x− + − =

1 0x− = 31 0x + =

⇒1 3 1 3

1, 1, ,2 2

i ix

+ −= −

1.−

1 0 0

5 2 0

1 6 1

A

= −

⇒

2 5 32 2 0 0

( ) 0 1 6 5 1 0

0 0 2 32 6 2

T

adj A

− = − = − −

0.234 0.2343434.....=

0.2 0.034 0.00034 0.0000034 ...+ + + +

34 34 340.2 .....

1000 100000 10000000+ + + + ∞

3 5 7

2 1 1 134 ........

10 10 10 10

= + + + + ∞

32 1/10 2 1 10034 34

10 1 1/1000 10 1000 99

= + = + × × −

2 34 232.

10 990 990= + =

2 3 5 5 2 5(1 ) (1 ) (1 )x x x x x+ + + = + +

2 3 4 5(1 5 10 10 5 )x x x x x= + + + + +

2 4 6 8 10(1 5 10 10 5 )x x x x x× + + + + +

5(1 10 5).2 16 32 512= + + = × =

52 2n = =

2(1 2 3 ) xx x e−− +2 3

2(1 2 3 ) 11! 2! 3!

x x xx x

= − + − + − +

K

5x

1 1 11 ( 2) 3

5 ! 4 ! 3!

= − + − + −

1 1 1 71.

120 12 2 120= − − − = −

1 2 3 4 5 6 7x x x x x x x

1x 1, 2, 3, ,9K

2 3 7, ,x x xK

K

1 2 6, , ,x x xK

1 2 6x x x+ + +K

7x K

9 .10 .10.10 .10.10 . 5 4500000.= =

( ) ( )P E P F E F≤ ⇒ ⊆

( ) 0P E F E F∩ > ⇒ ⊂

⇒

⇒

⇒

1 tan 1 tantan tan

4 4 1 tan 1 tan

π π θ θθ θ

θ θ

+ − + − − = − − +

2 2

4 tan 2 tan2 2 tan 2

1 tan 1 tan

θ θθ

θ θ

= = = − −

38°

2.05 m S

10°

O

Q P

l

Mathematics190

13. (c)

but and

Therefore,

Hence is continuous at

14. (a) ( )( )

( )( )

x a x by

x c x d

− −= − −

⇒ 1

log [log( ) log( ) log( ) log( )]2

y x a x b x c x d= − + − − − − −

Differentiating w.r.t. x we get

1 1 1 1 1 1

2 ( ) ( ) ( ) ( )

dy

y dx x a x b x c x d

= + − − − − − −

Thus 1 1 1 1

.2 ( ) ( ) ( ) ( )

dy y

dx x a x b x c x d

= + − − − − − −

15. (d)

∴

and as

Hence has minimum value –1 and also there is no

maximum value.

Alternate:

There is only one critical point having minima.

Hence has least value at

16. (a) Put then

17. (c) Required area

Trick: Area of sector made

by an arc

18. (b) ⇒ Now integrate it.

19. (a) Here, equation of line is tan , 0y x c cα= + >

Length of the perpendicular drawn on line from point

)sin,cos( αα aa

2

sin cos tan; cos

sec1 tan

a a c cp p c

α α αα

αα

− + += = =

+

20. (d) It is a fundamental concept.

21. (a)

Since, distance of vertex from origin is and focus is

V(1, 1) and F(2, 2) (ie, lying on y = x)

where, length of latusrectum

By definition of parabola

Where, PN is length of perpendicular upon

(ie, tangent at vertex).

22. (b) ,

23. (a) Here,

Now,

2

0

1lim ( ) sin ,x

f x xx+→

=

11 sin 1

x− ≤ ≤ 0x →

0 0lim ( ) 0 lim ( ) (0)x x

f x f x f+ −→ →

= = =

( )f x 0.x =

2 2

2 2 2

1 1 2 2( ) 1

1 1 1

x xf x

x x x

− + −= = = −

+ + +

( ) 1f x x< ∀

1≥ −2

22

1x≤

+

∴ 1 ( ) 1f x− ≤ <

( )f x

2 2

2 2 2 2

( 1)2 ( 1)2 4( )

( 1) ( 1)

x x x x xf x

x x

+ − −′ = =

+ +

( ) 0 0f x x′ = ⇒ =

2 2 2

2 4

( 1) 4 4 .2( 1)2( )

( 1)

x x x xf x

x

+ − +′′ =

+

2 2

2 3 2 3

( 1)4 16 ( ) 12 4

( 1) ( 1)

x x x x

x x

+ − − += =

+ +

∴ (0) 0f ′′ >

∴

( )f x 0.x =

min

1(0) 1.

1f f

−= = = −

log ,x x

ea t a a dx dt= ⇒ =

2 2

1

log1 1

x

xe

a dtdx

aa t=

− −∫ ∫

111 sin ( )

sin ( ) .log log

x

e e

at c c

a a

−−= + = +

3 22

0 34

3

xdx x dx= + −∫ ∫

3 222 1

30

1 44 sin

2 2 2 23

x x xx

− = + − +

3 3 2.

2 2 3 3

π ππ

= + − − =

2

2

c Rθ=

4. .

6 2 3

π π= =

2

2dy

dx x=

2

2,dy dx

x=

2

2 2.

∴

4a= 4 2= ( 2 )a =Q

∴ 2 (4 )( )PM a PN=

2 0x y+ − =

⇒2( ) 2

4 22 2

x y x y− + − =

⇒ 2( ) 8( 2)x y x y− = + −

| | | |a b a b⋅ = −r rr r

( cos 1)θ = −Q

cos , cos , cos ,l m nθ β θ= = = ( )l n=Q

2 2 21l m n+ + =

x 'x

'y

y

N

P

M

F (2,2)

y = x

(1,1) V

O

x + y – 2 = 0

yx 3=

(2,0) X

Y

191Mock Test-4

Given,

⇒

24. (b) .

25. (b) . . .(i)

. . .(ii)

and given . . .(iii)

Solving (i) and (iii), we get

Substituting these values in (ii), we get .

26. (a) Since , , ,a b c d are in H.P., therefore b is the H.M. of

a and c 2ac

ba c

=+

and c is the H.M. of b and d

i.e. 2

,bd

cb d

=+

∴2 2

( )( ) .ac bd

a c b db c

+ + =

⇒ 4ab ad bc cd ad+ + + = ⇒ 3 .ab bc cd ad+ + =

Trick : Check for .

27. (d) We have

28. (a) Total number of arrangements of word BANANA

The number of arrangements of words BANANA in

which two N’s appear adjacently

Required number of arrangements = 60 – 20 = 40

29. (a) Let1

1siny x−= and 1 2

2cos 1y x−= −

Differentiating w.r.t. x of 1y and

2 ,y we get

1

2

1

1

dy

dx x=

−

2 2

2 21

1 1( 2 ) 11.

2 11 (1 ) 1

dy dyx

dx dyxx x

−= − = ⇒ =

−− − −

30. (b) Let

⇒

Hence for every therefore it is non-

decreasing in [0, 2].

JEE Advance Paper-I

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

4 5 3 4 2 9 2 3 a b,c,d

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

a b b c a,b c b,d d b a

1. (4)

2. (5) Coeff. 2x

⇒ 2 3 4 49 50 2 51

2 2 2 2 2 3....... (3 1)C C C C C m n C+ + + + + = +

⇒ 3 3 4 49 50 2 51

3 2 2 2 2 3....... (3 1)C C C C C m n C+ + + + + = +

⇒ 1

1

n n n

r r rC C C+−+ = ⇒ 50 50 2 51

3 2 3(3 1)C C m n C+ ⋅ = +

⇒ 50 50 2 50 50 51

3 2 2 2 3

51( 1) 3

3C C m C n C C+ + − = ⋅ +

⇒ 51 2 50 50 51

3 2 2 3( 1) 51C m C n C C+ − = ⋅ +

⇒ 2 1 51m n− = 2 51 1m n= +

Min value of 2m for 51 1n+ is integer for 5.n =

3. (8) Let coin was tossed ‘n’ times

Probability of getting at least two heads

4. (4)

f (x) is periodic with period 2

put

⇒ 2 22cos cos 1θ β+ =

⇒ 2 22cos sinθ β= 2 2sin 3sinβ θ=

⇒ 2 22cos 3sinθ θ= 2

5cos 3,θ =

∴ 2 3cos .

5θ =

qpqp ∧=∧))(~(~

6α β+ = −

αβ λ=

3 2 20α β+ = −

2, 8β α= = −

16λ = −

..ei

4

1,

3

1,

2

1,1 ==== dcba

2 33 6 10 ....y x x x= + + +

∴ 2 31 1 3 6 10 ...y x x x+ = + + + +

⇒ 3 1/31 (1 ) 1 (1 )y x x y− −+ = − ⇒ − = + ⇒ 1/31 (1 )x y −= − +

6!60

3!2!= =

5!20

3!= =

2( ) xy f x x e−= =

2 22 (2 )x x xdyxe x e e x x

dx

− − −= − = −

'( ) 0f x ≥ [0, 2],x ∈

127 7

1

3(4 2) 7

1

112

43

1

π π

π

=

−

=

Σ −

= =

Σ −

ki i

k

ii k

k

e e

e e

11

2 2n n

n = − +

⇒1

1 0.962

n

n + − ≥

⇒2

251

n

n≥

+⇒ 8n ≥

1, 1 2( )

1 , 0 1

x xf x

x x

− ≤ <=

− ≤ <

∴10

10

( ) cosI f x x dxπ−

= ∫10 2

0 0

2 ( ) cos 2 5 ( ) cosf x x dx f x x dxπ π= = ×∫ ∫1 2

1 2

0 1

10 (1 )cos ( 1)cos 10( )x xdx x x dx I Iπ π

= − + − = + ∫ ∫

2

2

1

( 1)cosI x x dxπ= −∫ 1x t− =

1

1 – 2 – 1 2 3

Mathematics192

5. (2)

6. (9) 1 1( )y y m x x− = −

Put 0,x = to get y intercept 3

1 1 1y mx x= =

3

1 1 1

dyy x x

dx− =

3dy

x y xdx

− = −

2dy y

xdx x

− = −

1

ln 1f dxxxe e

x

−= =

21 1

y x dxx x

× = − ×∫

2

2

y y xxdx c

x x= − ⇒ = − +∫

⇒

3 3( ) ( 3) 9.

2 2

xf x x f= − + ∴ − =

7. (2) Substituting in

Also,

Also,

8. (3) 2 2 2

9a b b c c a− + − + − =r rr r r r

⇒ 2 2 2 2 2 2 9a b b c c a− ⋅ + − ⋅ + − ⋅ =r rr r r r

⇒ 3

2a b b c c a− = ⋅ + ⋅ + ⋅

r rr r r r . . .(i)

Now, 2| | 0a b c+ + ≥

rr r

⇒ 1 1 1 2 0a b b c c a+ + + ⋅ + ⋅ + ⋅ ≥r rr r r r

⇒ 3

2a b b c c a⋅ + ⋅ + ⋅ ≥ −

r rr r r r . . .(ii)

Equation (i) and (ii) are simultaneously true

If 1

2a b b c c a⋅ = ⋅ = ⋅ ≥ −

r rr r r r

Now, 2

2 5 5a b c+ +rr r

4 25 25 20 50 20( )a b b c a c= + + + ⋅ + ⋅ + ⋅r rr r r r

54 10 25 10 9= − − − = ⇒ 2 5 5 3a b c+ + =rr r

9 (a)

10. (b,c,d) For

also

is decreasing for

Also,

1

2

0

cosI t t dtπ= −∫1 1

1

0 0

(1 )cos cos( )I x x dx x x dxπ π= − − = −∫ ∫

∴1

0

10 2 cosI x x dxπ

= −

∫1

2

0

sin cos20

x xx

π ππ π

= − +

2

2 2 2

1 1 4020 4

10I

ππ π π

= − − − = ∴ =

1 23 2 5

2I II0

4 (1 )−∫d

x x dxdx

1 1

3 2 5 2 2 5

0 0

4 (1 ) 12 (1 ) = − − − ∫

d dx x x x dx

dx dx

11 1

3 2 4 2 2 5 2 5

0 00

4 5(1 ) ( 2 ) 12 (1 ) 2 (1 )

= × − − − − − −

∫x x x x x x x dx

1

2 5

0

0 0 12[0 0] 12 2 (1 )= − − − + −∫ x x dx

12 6

0

(1 )12

6

−= × −

x

112 0 2

6

= + =

,0a

e

2 1y x= − +

20 1

a

e= − +

21

a

e=

2

ea =

2 2 21 a m b= −

2 21 a m b2= −2 21 4a b= −

224

14

eb= −

2 2 1.b e= = 2 2 1( 1)b a e= −

∴ 1, 2a e= =

1sec(tan )y x−=

⇒ 1 1

2

1sec(tan ).tan (tan ).

1

dyx x

dx x

− −=+

⇒1

2 1

1 1 2x

dy

dx =

= = +

1f (x) x cos ,x 1

x

= ≥

1 1 1f '(x) cos sin 1forx

x x x

= + → → ∞

2 2 3

1 1 1 1 1 1f '(x) sin sin cos

x x x x x x

= − −

3

1 1cos 0 forx 1

x x

= < ≥

⇒ f '(x) [1, )∞

⇒ f '(x 2) f '(x).+ <

x x

1 1lim f (x 2) f (x) lim (x 2)cos x cos 2

x 2 x→∞ →∞

+ − = + − = +

∴ f (x 2) f (x) 2 x 1+ − > ∀ ≥

193Mock Test-4

11. (a) Here 1/ 1/

( )( )

[1 ( ) ] (1 2 )= =

+ +n n n x

f x xff x

f x x

⇒ 1/

( )(1 3 )

=+ n n

xfff x

x

⇒ 1/

times

( ) ( )( )(1 )n n

n

xg x f f f x

nx= =

+o oKo

Hence

12

1/( )

(1 )

−−= =

+∫ ∫n

n

n n

x dxI x g x dx

nx

2 1

2 1/ 2 1/

(1 )1 1

(1 ) (1 )

− += =

+ +∫ ∫n

n

n n n n

dnx

n x dx dx dxn nx n nx

∴

111

(1 )( 1)

−= + +

−n nI nx K

n n

12. (b)

13. (b)

This is a linear differential equation

I.F. =

Solution is

or

Now, (Using property)

=

(Taking )

14. (c) Since, ∆ is isosceles, hence centroid is the desired

point.

15. (a,b) Slope of tangent = m = 2 Equation of tangent in

slope form is

and point of contact is

16. (c)

Area of base (PQRS)

Height = proj. of PT on

Volume

17. (b, d) Let the required plane be

Distance of from is 2

18. (d)

(0) 0F =

( ) 2 ( ) ( )F x xf x f x′ ′= =2

( ) x cf x e +=2

( ) ( (0) 1)xf x e f= =Q

2

0

( )

x

xF x e dx= ∫

2

( ) 1xF x e= − ( (0) 0)F =Q ⇒ 4(2) 1F e= −

4

2 2

2

1 1

dy x x xy

dx x x

++ =

− −

22

1ln| 1|

21 2 1

xdx x

xe e x−

−∫ = = −

⇒3

2 2

2

( 2)1 1

1

x xy x x dx

x

+− = ⋅ −

−∫

52 4 2

1 ( 2 )5

xy x x x dx x c− = + = + +∫

(0) 0f = ⇒ 0c =

⇒5

2 2( ) 1

5

xf x x x− = +

3 / 2 3 / 2 2

23 / 2 3 / 2

( )1

xf x dx dx

x− −

=−

∫ ∫3 / 2 / 32 2

20 0

sin2 2 cos

cos1

xdx d

x

π θθ θ

θ=

−∫ ∫

sinx θ=/ 3/ 3

2

00

sin 22 sin

2 4d

ππ θ θθ θ = = − ∫

3 32 2 .

6 8 3 4

π π = − = −

2 2 2, 2 4 2y mx a m b y x= ± − = ±

2 2

,ma b

c c

−−

2 9 4 9 1, ,

4 2 4 2 2 2 2

× ≡ − − ≡ ± ± ± ±

ˆˆ ˆ1 1

| | 3 1 22 2

1 3 4

i j k

PR SQ= × = −

− −

uuur uuur

1 ˆ ˆˆ ˆ ˆ ˆ| 10 10 10 | 5 | | 5 32

i j k i j k= − + − = − + =

1 2 3 2ˆˆ ˆ3 3

i j k− +

− + = =

2(5 3) 10 cu. units

3

= =

1 0x z yλ+ + − =

⇒2

| 1 | 11

22

λλ

λ

−= ⇒ = −

+

⇒ 3 2 2 2 0P x y z≡ − + − =

3P ( , , )α β γ| 2 2 2 |

24 1 4

α β γ− + −=

× +

⇒ 2 2 4 0 and 2 2 8 0α β λ α β λ− + + = − + − =

~ [~ (~ )]s r s∨ ∧

~ (~ ) ~ (~ )s r s= ∧ ∧

( ~ )s r s= ∧ ∨

( ) ( ~ )s r s s= ∧ ∨ ∧

s r F= ∧ ∨ s r= ∧

S P

QR

T( , , )α β γ

(6, 0)

(3, 4)

(0, 0)

Mathematics194

19. (A) 2 2 2

2 ( ) 2 ( )

1 ( ) 1 ( 2 )

+ += =

− + − − +i x iy i x iy

zx iy x y ixy

Using 2 21− =x y

2

2 2 1.

2 2

−= = −

−ix y

Zy ixy y

Q 1 1

1 1 1 1.− ≤ ≤ ⇒ − ≤ − − ≥y ory y

(B) For domain 2

2( 1)

8.31 1

1 3

−

−− ≤ ≤−

x

x

⇒ 2

2 2

3 31 1.

1 3

−

−

−− ≤ ≤

−

x x

x

Case (i): 2

2 2

3 31 0

1 3

−

−

−− ≤

−

x x

x

⇒ 2

2 2

(3 1)(3 1)0

(3 1)

−

−

− −≥

−

x x

x

⇒ ( ,0] (1, ).∈ −∞ ∪ ∞x

Case (ii): 2

2

3 31 0

1 3 2

−−+ ≥

− −

x x

x

⇒ 2

2

(3 1)(3 1)0

(3 .3 1)

−

−

− +≥

−

x x

x x

⇒ ( , 1) [2, ).∈ −∞ ∪ ∞x

So, ( , 0] [2, )∈ −∞ ∪ ∞x

(C) 1 1 3→ +R R R

0 0 2

( ) tan 1 tan

1 tan 1

θ θ θθ

= −

− −

f 2 22(tan 1) 2sec .θ θ= + =

(D) 1/ 2 3/ 2 1/23 15

( ) ( ) (3 10) ( ) 3 ( ) ( 2)2 2

′ = − + × = −f x x x x x x

Increasing, when 2.≥x

20. . The graph of is shown

(A) If

(B) If

(C) If

(D) If


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

4824 1004 50 3 40 30 1601 500 c d

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

c a,d b c a,c,d c a b b c

1. (4824) The period of is as

Similarly the period of is 3 and the period of

is

Hence, the period of the given function

LCM of 24, 8, 3 = 24

2. (1004)

3. (50)

4. (3) Let

( 1)( 5)( )

( 2)( 3)

x xf x

x x

− −=

− −( )f x

1 1x− < < ⇒ 0 ( ) 1f x< <

1 2x< < ⇒ ( ) 0f x <

3 5x< < ⇒ ( ) 0f x <

5x > ⇒ ( ) 1f x <

[ ]sin

12

xπ

224

/12

ππ

=

[ 24] ([ ] 24)sin sin

12 12

x xπ π+ + =

[ ] [ ]sin 2 sin

12 12

x xπ ππ = + =

[ ]tan

3

xπ

cos4

xπ

28

/ 4

ππ

=

λ =

∴ 201 201 24 4824λ = × =

Q 1 1tan tan

1

bx

ax b aybbx a

xa

− −

− − = = + + ⋅

1 1tan tan

bx

a

− − = −

∴2

10

1

dy

dx x= −

+

∴2

1

1 1 1

1 ( 1) 1 1 2x

dy

dx =−

= = =+ − +

⇒1

1(2008) 2008 1004

2x

dy

dx =−

= × =

2

4

33000 33000

PL d NPLAN

I

π = =

∴2I P L d N

I P L d N

∆ ∆ ∆ ∆ ∆= + + +

⇒ 100 100 100 2 100 100I P L d N

I P L d N

∆ ∆ ∆ ∆ ∆ × = × + × + × + ×

10% 10% 2 10% 10% 50%= + + × + =

∴ 50λ =

1

2

2 2sin

(4 8 13)

xI dx

x x

− + = + +

∫

y

x 1 2 3 5

y =

195Mock Test-4

Put

Then

Hence

Then,

5. (40)

Putting we get

For

6. (30) Since, centroid divides the orthocenter and circum

center in the ratio (internally) and if centroid

then

and

Centroid is and nine point centre is the mid point

of orthocenter and circumcentre.

∴ Nine point centre is

i.e.,

and

7. (1601) Solving the equation and

then,

∴

or

Therefore the centre of the required circle is

but circle passes through (2,0)

Radius of the required circle

8. (500) Let PV of A, B, C and D be and

and

Adding all we get

(Area of )

Then,

9. (c)

Having elements

Total subsets of is

Total no. of subsets of having 3 or more elements

1

2 2

2 2sin

(2 2) 3

xI dx

x

− + = + +

∫

2 2 3tanx θ+ =

∴ 22 3secdx dθ θ=

23sec 3 tan lnsec

2 2

dI c

θ θθ θ θ θ= ⋅ = ⋅ − +∫

2

13 2 2 2 2 2 2tan ln 1

2 3 3 3

x x xc−

+ + + = ⋅ − + +

1 22 2 3( 1) tan ln(4 8 13)

3 4

xx x x c− + = + − + + +

3

4λ = −

4 3λ− =

1dx

xdt

= +

⇒1

dxdt

x=

+

⇒ ln( 1)x t c+ = +

0, 0t x= = 0c =

⇒ ln( 1)t x= +

99, ln100 2log 10e

x t= = =

∴10 10

20 log 20 2log 10 log 40e

e eλ = × × =

2 : 1

( , ),G x y2 1

13 2

(1, ) ( , ) ,2 4

O G x y C

32 1 1

42

2 1 3x

× + ×= =

+

32 1 1

54

2 1 6y

× + ×= =

+

∴4 5

,3 6

( ) ( )1 3/ 2 1 3/ 4, ,

2 2

+ +

5 7,

4 8

⇒4 5

,3 6

a β= =5 7

,4 8

γ δ= =

∴ 6 12 4 8α β γ δ+ + +

4 5 5 76 12 4 8 8 10 5 7 30

3 6 4 8= × + × + × + × = + + + =

2(2 ) 5 1c x c y+ + =

3 5 1,x y+ =

2 1 3(2 ) 5 1,

5

xc x c

− + + =

2(2 ) (1 3 ) 1c x c x+ + − =

2

2

1

2 3

cx

c c

−=

+ −(1 )(1 ) 1

(3 2)(1 ) 3 2

c c cx

c c c

+ − += =

+ − +

∴1

1lim

3 2c

cx

c→

+=

+⇒

2

5x =

∴ 1 3 1 (6 / 5) 1

5 5 25

xy

− −= = = −

2 1,

5 25

−

∴2 2

2 12 0

5 25

= − + − −

64 1 1601 1601

25 625 625 25 25

λ= + = = =

∴ 1601λ =

, ,a b crr r

0r

∴ ( )AB CD b a c b c a c× = − × − = − × + ×uuur uuur r rr r r r r

( )BC AD c b a c a b a× = − × − = − × + ×uuur uuur r rr r r r r

( )CA BD a c b a b c b× = − × − = − × + ×uuur uuur r r rr r r r

2( )AB CD BC AD CA BD a b b c c a× + × + × = − × + × + ×uuur uuur uuur uuur uuur uuur r rr r r r

∴ | | 2 | |AB CD BC AD CA BD a b b c c a× + × + × = × + × + ×uuur uuur uuur uuur uuur uuur r rr r r r

2 | ( ) ( ) |c a b a= − × −rr r r

2 | |AC AB= ×uuur uuur

14 | | 4

2AC AB= ⋅ × =uuur uuur

ABC∆

∴ 4λ =

125 500λ =

[ , ]A x y= , , , dB a b c=

A B× 2 4 8× =

A B× 82 256=

∴ A B×

8

2null set single ton set subset having

2 elements

256 1 8 C

= − + +

256 1 5 28 219= − − − =

Mathematics196

10. (d)

Not possible. As condition for two distinct real root is

(where are roots of

11. (c)

But

Hence, z lies on a circle of radius 2 centered at origin.

12. (a,d)

1 4 4

2 1 7 | | 4

1 1 3

adj P adj P

= ⇒ =

We know, 1| | | |nadj P P −= (where n is order of matrix)

⇒ 2| | | |adj P P= ⇒ 24 | |P= ⇒ | | 2P = ±

13. (b)

2 2 2 2 28 12 16 20 24

...5 5 5 5 5

+ + + + +

2 2 2 2 2

2 2 2 2 2

8 12 16 20 24....

5 5 5 5 5= + + + + +

14. (c) Number of required ways

=

15. (a,c,d,) Conditions:

tan( ) 0 tan 0θ θ− > ⇒ < and 3

1 sin2

θ− < < −

∴ 3 5 1

, 0 cos2 3 2

π πθ θ ∈ ⇒ < <

Also, 2 12cos (1 sin ) sin cos 1

sin cos2 2

θ φ θ φθ θ

− = −

⇒ 2cos 2cos sin 2sin cos 1θ θ φ θ φ− = −

⇒ 1 2cos 2sin( )θ θ φ+ = +

⇒ 1

sin( ) cos2

θ φ θ+ = +

⇒ 1 4

sin( ) 12 2 3

π πθ φ φ< + < ⇒ < <

16. (c)

17. (a)

is following Fibonacci series. Hence

18. (b)

Total no of ways exactly two consecutive two is = 5 ways

Total no of ways exactly three consecutive two is = 2

ways

Total no of ways exactly four consecutive two is = 1 ways

So,

19. (b) Case (i): One odd, 2 even

Total number of ways = 2×2×3+1×3×3+1×2×4 = 29.

Case (ii): AII 3 odd

Number of ways = 2×3×4 = 24

Favourable ways = 53

Required probability =

20. (c) Here

even

Hence number of favourable ways

3( ) 2 3f x x x k= + +2( ) 6 3f x x′ = + ( ) 0f x′ =

⇒ 2 1

2x = −

( ) ( ) 0f fα β = ,α β ( ) 0)f x′ =

1 2

1 2

21

2

−=

−z z

z z⇒ 2 2

1 2 1 2| 2 | | 2 |− = −z z z z

⇒ 1 2 1 2 1 2 1 2( 2 )( 2 ) (2 )(2 )− − = − −z z z z z z z z

⇒ 1 1 2 2 1 1 2 24 4+ = +z z z z z z z z

⇒ 2 2 2 2

1 2 2 14 | | | | 4 | | | | 0z z z z+ − − =

⇒ 2 2

1 2(| | 1) ( | | 4) 0z z− ⋅ − =

2| | 1,z ≠ ∴2| | 2z =

2

2

(4 4)

5

+=

n

nT

10 102 2

21 1

1 1616( 1) ( 2 1)

255n

n n

S n n n= =

= + = + +∑ ∑

16 10 11 21 2 10 11 16 1610 505

25 6 2 25 5

× × × × = + + = × = m

⇒ 101=m

4 4

2 35! 4 4! 3! 2! 1 53.C C− ⋅ − ⋅ + ⋅ − =

1/ 22 1 1 21 ( cos cot sin cot ) 1x x x x− − + + −

1/ 22

2 1 1

2 2

11 cos cos sin sin 1

1 1

xx x

x x

− − = + + − + +

1/ 22

22

2 2

11 1

1 1

xx

x x

= + + − + +

2 2 1/ 2 21 ( 1 1) 1 .x x x x= + + − = +

11a =

22a =

33a =

45a =

na

17 16 15.a a a= +

1 1

65 2 1b = + + 8=

53 53.

3 5 7 105=

× ×

2 1 32x x x= +

⇒1 3x x+ =

2 4 1 1 3

1 2 1 11.C C C C= ⋅ + ⋅ =

197Mock Test-5


1. The domain of definition of the function given by

the equation is:

a. b.

c. d.

2. A real root of the equation is

a. 1 b. 2 c. 3 d. 4

3. is equal to

a. b.

c. d.

4. Matrix is

a. Orthogonal b. Idempotent

c. Skew-symmetric d. Symmetric

5. For a sequence and Then is

a. b.

c. d. None of these

6. Coefficients of in the expansion of

a. b.

c. d. None of these

7.

a. e b. c. e/2 d. e/3

8. The number of ways in which the following prizes be

given to a class of 20 boys, first and second Mathematics,

first and second Physics, first Chemistry and first English

is.

a. b.

c. d. None of these

9. If the integers m and n are chosen at random between 1

and 100, then the probability that a number of the form

is divisible by 5, equals

a. b. c. d.

10.

a. 1 b. c. d.

11. equals

a. b.

c. d.

12. A tower subtends an angle α at a point A in the plane of

its base and the angle of depression of the foot of the

tower at a point l meters just above A is .β The height of

the tower is

a. tan cotl β α b. tan cotl α β

c. tan tanl α β d. cot cotl α β

13. The value of k so that the function

is continuous at is

a. 1 b. 2


14. If (1 ) ,xy x= + then dy

dx=

a. (1 ) log1

x xx ex

x

+ + +

b. log(1 )1

xx

x+ +

+

c. (1 ) log(1 )1

x xx x

x

+ + + +

d. None of these

15. 1 tan

x

x x+ is maxima at

a. sinx x= b. cosx x=

c. 3

xπ

= d. tanx x=

16.

a. b.

c. d.

17. The area of the triangle formed by the tangent to the

hyperbola and co-ordinate axes is

a. b. c. d.

( )y x

2 2 2x y+ =

0 1x< ≤ 1x x≤ ≤

0x−∞ < ≤ 1.x−∞ < <

4 2log log ( 8 ) 0x x+ − =

cosh( ) cosh( )i iα β α β+ − −

2 sinh sinhα β 2 cosh coshα β

2 sinh sini α β 2 cosh cosα β

0 4 1

4 0 5

1 5 0

− − −

1, 2

na a< > = 1 1

.3

n

n

a

a

+ =20

1

r

r

a=∑

20[4 19 3]

2+ ×

20

13 1

3

−

202(1 3 )−

[0 ( 1)]rx r n≤ ≤ −1 2( 3) ( 3) ( 2)n nx x x− −+ + + +

3 2 1( 3) ( 2) ... ( 2)n nx x x− −+ + + + + +

(3 2 )n r n

rC − (3 2 )n n r n r

rC − −−

(3 2 )n r n r

rC −+

2 3 41

3! 5! 7!+ + + + ∞ =K

2e

4 220 19× 3 3

20 19×2 420 19×

7 7m n+

1

4

1

7

1

8

1

49

2sin( ) sin( ) cosecπ θ π θ θ+ − =

1– sinθ sinθ−

2sinh x

cosh 2 1x − 2cosh 1x +

1(cosh 2 1)

2x −

1(cosh 2 1)

2x +

2(2 ), when 0( )

cos , when 0

k x x xf x

x x

− <=

≥0,x =

1

1 logdx

x x=

+∫

3/ 22(1 log )

3x c+ + 3/ 2(1 log )x c+ +

2 1 log x c+ + 1 log x c+ +

2xy a=2a 22a 23a 24a

Mathematics198

18. The solution of is

a.

b.

c.

d. None of these

19. The distance of point (–2, 3) from the line is

a.

b.

c.

d.

20. If the coordinates of one end of the diameter of the circle

2 2 8 4 0x y x y c+ − − + = are (–3, 2), then the coordinates

of other end are

a. (5, 3) b. (6, 2)

c. (1, –8) d. (11, 2)

21. Let by any point on the parabola Let P be

the point that divides the line segment from to

in the ratio Then, the locus of P is

a.

b.

c.

d.

22. If are unit vectors such that then

a. 1

b. 3

c. – 3/2

d. 3/2

23. The direction cosines of three lines passing through the

origin are and The lines will

be coplanar, if

a. b.

c.

d. None of these

24. is equal to

a.

b.

c.

d.

25. If the sum of the roots of the equation

be equal to their product, then

a. 4

b. –4

c. 6

d. None of these

26. If and are the complex numbers representing

the vertices of two triangles such that and

where r is a complex number, than the

two triangles

a. have the same area

b. are similar

c. are congruent

d. None of these

27. The rank of the matrix, is

a. 2 b. 3

c. 1 d. Indeterminate

28. The coefficient of in the expansion of

is

a. 1

b. (–1)n

c. n

d. n + 1

29. Three boys and two girls stand in a queue. The probability

that the number of boys ahead of every girl is atleast one

more than the number of girls ahead of her, is

a. 1/2

b. 1/3

c. 2/3

d. 34

30. The imaginary part of is

a.

b.

c.

d. None of these

logdy

x xdx

=

22 log

2

xy x x c= − +

22log

2

xy x x c= − +

2 21 1log

2 2y x x x c= + +

5− =x y

5 2

2 5

3 5

5 3

( , )x y 2 4 .y x=

(0,0)

( , )x y 1: 3.

2x y=

2 2y x=

2y x=

2 2x y=

, ,a b crr r

0,a b c+ + =rr r

a b b c c a⋅ + ⋅ + ⋅ =r rr r r r

1 1 1 2 2 2, , ; , ,l m n l m n

3 3 3, , .l m n

1 1 1

2 2 2

3 3 3

0

l n m

l n m

l n m

=1 2 3

2 3 1

3 1 2

0

l m n

l m n

l m n

=

1 2 3 1 2 3 1 2 30l l l m m m n n n+ + =

~ ( (~ ))∨p q

~ ∨p q

(~ )∧p q

~ ~∨p p

~ ~∧p q

2 2 3 0x xλ λ+ + =

λ =

, ,a b c , ,u v w

(1 )c r a rb= − +

(1 ) ,w r u rv= − +

2 3 1 4

0 1 2 1

0 2 4 2

A

= − − −

nx

2(1 ....) nx x −+ + +

)sin(costan 1 θθ i+−

)(sintanh 1 θ−

)(tanh 1 ∞−

)(sintanh2

1 1 θ−

199Mock Test-5

JEE ADVANCE PAPER-I



9 (both inclusive).

1. A variable plane is at a constant distance p form the origin

and meets the axes in A, B and C. If the locus of the

centroid of the tetrahedron OABC is


2. The normal to the parabola at the point (2, 4)

meets it again at (18, –12). If length of normal chord is


3. An electric component manufactured by ‘RASU

electronics’ is tested for its defectiveness by a

sophisticated testing device. Let A denote the event “the

device is defective” and B the event “the testing device

reveals the component to be defective”. Suppose

and where < 1.

If the probability that the component is not defective,

given that the testing device reveals it to be defective, is

then the value of 2008 must be

4. The coefficient of in the polynomials after parenthesis

have been removed and like terms have been collected in

the expansion

is then the value of

must be

5. The value of

must be

6. Let and be three non-coplanar unit vectors such

that the angle between every pair of them is If

where and are

scalars, then the value of is _______

7. If then the value of

must be

8. If then the value

of must be



9. Let A and B be two sets containing four and two elements

respectively. Then the number of subsets of the set

each having at least three elements is:

a. 219 b. 256

c. 275 d. 510

10. The quadratic equation with real coefficients

has purely imaginary roots. Then the equation

has

a. only purely imaginary roots

b. all real roots

c. two real and two purely imaginary roots

d. neither real nor purely imaginary roots

11. A value of for which is purely imaginary, is:

a. b.

c. d.

12. Let be a complex cube root of unity with and

be a matrix with Then when

a. 57 b. 55

c. 58 d. 56

13. If m is the A.M. of two distinct real numbers and

and G1, G2 and G3 are three geometric means

between and n, then equals,

a. b.

c. d.

2 2 2 2x y z pλ− − − −+ + = 160λ

2 8y x=

,λ2λ

( )P A α= ( / ) ( '/ ') 1 ,P B A P B A α= = − 0 1.α< <

,λ λ

50x

1000(1 )x+ + 999(1 )x x+ 2x+ 998 1000(1 ) ...x x+ + +

998 1000(1 ) ...x x+ + +!

,! !v

λµ

2λ µ+ 3v+

( )v µ>

,a brr

cr

.3

π

,a b b c pa qb rc× + × = + +r r rr r r r

,p q r

2 2 2

2

2p q r

q

+ +

2 22 22 , 2 ,

8 8

y yf x x xy

+ − =

(60, 48) (80, 48) (13, 5)f f f+ +

(cos sin ); (sin cos ),x a t t t y a t t t= + = −

2

2

/3

120)

t

d y

dxπ=

,×A B

( ) 0=p x

( ( )) 0−p p x

θ 2 3 sin

1 2 sin

i

i

θθ

+−

3

π6

π

1 3sin

4

−

1 1sin

3

−

ω 1ω ≠

[ ]ijP P ×n n .ω += i j

ijp

2 0,≠P =n

l

( , 1)n n >l

l4 4 4

1 2 32G G G+ +24 mnl

24 m nl

24 mnl2 2 24 m nl

1 1 1 11 7 2 5216 sin sin 27cos cos 28 tan tan 200cot cos

6 3 4 4

π π π π− − − − − + + +

1 1 1 11 7 2 5216 sin sin 27cos cos 28 tan tan 200cot cos

6 3 4 4

π π π ππ

− − − − − + + +

Mathematics200

14. Let Then can take value(s)

a. 1056 b. 1088 c. 1120 d. 1332

15. If then

a. b.

c. d.

16. The number of integers greater than 6,000 that can be

formed, using the digits 3, 5, 6, 7 and 8, without

repetition, is:

a. 216 b. 192 c. 120 d. 72

17. A ship is fitted with three engines and The

engines function independently of each other with

respective probabilities and For the ship to be

operational at least two of its engines must function. Let X

denote the event that the ship is operational and let

and denote respectively the events that the

engines And are functioning. Which of the

following is (are) true?

a.

b. P [Exactly two engines of the ship are functioning

c.

d.

18. Let PQR be a triangle of area with and

where a, b and c are the lengths of the sides of the

triangle opposite to the angles at P, Q and R respectively.

Then equals

a. b. c. d.



Column II.

19. Match the statements given in Column I with the

interval/union of intervals given in Column II

Column I Column II

(A) The set is a

complex number,

is

1.

(B) The domain of the function

is

2.

(C) If

then the set

is

3.

(D) If

then f(x) is

increasing in

4.

5.

20. Match the statements/expressions in Column I with the

values given in Column II.

Column I Column II

(A) The number of solutions of the

equation sin cos 0xxe x− = in the

interval 0,2

π

1. 1

(B) Value (s) of k for which the plane

4 0,4 2 0kx y z x ky z+ + = + + = and

2 2 0x y z+ + = intersect in a

straight line

2. 2

(C) Value (s) of k for which

| 1 | | 2 | | 1 | 2 | 4x x x x k− + − + + + =

has integer solution (s)

3. 3

(D) If 1y y′ + and (0) 1y = then value

(s) of y (ln2)

4. 4

5. 5

( 1)422

1

( 1) .k kn

n

k

S k+

=

= −∑ nS

13 4 ,−=x x =x

3

3

2 log 2

2 log 2 1− 2

2

2 log 3−

4

1

1 log 3−2

2

2 log 3

2 log 3 1−

1 2,E E 3.E

1 1,

2 4

1.

4

1 2,X X 3X

1 2,E E 3E

1

3[ | ]

16

cP X X =

7[ ]

8X =

2

5[ | ]

16P X X =

1

7[ | ]

16P X X =

∆7

2,2

a b= =

5,

2c =

2sin sin 2

2sin sin 2

P P

P P

−

+

3

4∆45

4∆

23

4

∆

245

4

∆

2

2izRe : z

1 z

−

| z | 1, z 1= ≠ ±

( , 1) (1, )−∞ − ∪ ∞

1f (x) sin−=x 2

2(x 1)

8(3)

1 3

−

−

−

( , 0) (0, )−∞ ∪ ∞

1 tan 1

f ( ) tan 1 tan ,

1 tan 1

θθ θ θ

θ= −

− −

f ( ) : 02

πθ θ ≤ <

[2, )∞

3 / 2f (x) x (3x 10),= −

x 0,≥

( , 1] [1, )−∞ − ∪ ∞

( ,0] [2, )−∞ ∪ ∞


201Mock Test-5




9 (both inclusive).

1. If the approximate value of 0 abcdef, It is

given that and then the

value of abcd must be

2. If then the

value of must be

3. The solution of the differential equation

satisfying 1

(0) ,8

y = y1 (0) = 0and y2 (0) = 1is

then the numerical value of λ must be

4. If and are the roots of the equation

and if area of the triangle

formed by the lines and is

then the value of 2008 must be

5. Tangents are drawn form to the circle

then the radius of the circle such that the area of the ∆ formed

by tangents and chore of contact is maximum must be

6. ‘P’ is any arbitrary point on the circum circle of the

equilateral triangle of side length 26 unit, then the value

of must be

7. The lines and

are coplanar for k is equal to

8. If a circle cuts a rectangular hyperbola in A, B, C

and D are the parameters of these four points be

and respectively, then the value of must be



9. Let where

Then value of is

a. b.

c. d.

10. If the angles of elevation of the top of a tower from three

collinear points A, B and C, on a line leading to the foot

of the lower, are and respectively, then the

ratio, AB : BC is

a. b.

c. d. 2 : 3

11. Let and are

integers, and let be the left hand

derivative of at If then

a. b.

c. d.

12. Let where is a twice differentiable

positive function on such that

Then, for

a.

b.

c.

d.

13. Let Then for

an arbitrary constant the value of equals

a.

10log (4.04)

4log 4 0.6021=

10log 0.4343,e =

1 tan 1( tan cot ) tan ,

tan

xx x dx a c

b x

− − + = +

∫

4 5a b+

3 2

3 28 0

d y d y

dx dx− =

8

1

8 7,

xe xy

λ− +

=

1m

2m

2 ( 3 2) ( 3 1) 0x x+ + + − =

1 2 2, ,y m x y m x= = y c=

2( ) ,a b c+ 2 2

( )a b+

(6, 8)P2 2 2 ,x y r+ =

2 2 2| | | | | |PA PB PC+ +uuur uuur uuur

4 6 1

3 5 2

x y z+ + −= =

−3 2 5 0x y z− + + =

2 3 4x y z k= + + −

2xy c=

1 2 3, ,t t t

4t 1 2 3 416t t t t

1 1 1

2

2tan tan tan ,

1

xy x

x

− − − = + −

1| | .

3x <

a y

3

2

3

1 3

x x

x

−−

3

2

3

1 3

x x

x

+−

3

2

3

1 3

x x

x

−+

3

2

3

1 3

x x

x

++

30 , 45° ° 60°

3 :1 3 : 2

1: 3

m

( 1)( ) ;

log cos ( 1)

nxg x

x

−=

−0 2,x< < m n

0, 0,≠ >m n p

| 1|x− 1.x =1

lim ( ) ,x

g x p+→

=

1, 1n m= = 1, 1n m= = −

2, 2n m= = 2,> =n m n

( ) log( ( ))=g x f x ( )f x

(0, )∞ ( 1) ( ).+ =f x x f x

1, 2, 3, ...,=N

2

1 1 14 1 ...

9 25 (2 1)

− + + + +

− N

2

1 1 14 1 ...

9 25 (2 1)

+ + + +

− N

2

1 1 14 1 ...

9 25 (2 1)

− + + + +

+ N

2

1 1 14 1 ...

9 25 (2 1)

+ + + +

+ N

4 2 4 2, .

1 1

−

− −= =+ + + +∫ ∫

x x

x x x x

e eI dx J dx

e e e e

,C −J I

4 2

4 2

1 1log

2 1

− ++ + +

x x

x x

e eC

e e

Mathematics202

b.

c.

d.

14. The integral equals:

a.

b.

c.

d.

15. Let the population of rabbits surviving at a time t be

government by the differential equation

If then (p)t equals:

a.

b.

c.

d.

16. A straight line L through the point (3, –2) is inclined at an

angle to the line If L also intersects the

x-axis, then the equation of L is

a.

b.

c.

d.






Consider the line

17. The unit vector perpendicular to both and is

a.

b.

c.

d.

18. The distance of the point (1, 1, 1) from the plane passing

through the point (−1, −2, −1) and whose normal is

perpendicular to both the lines and is

a.

b.

c.

d.


Let ABCD be a square of side length 2 units, is the circle

through vertices, A, B, C, D and is the circle touching all the

sides of the square ABCD. L is a line through A.

19. If P is a point on and Q in another point on then

is equal to

a. 0.75

b. 1.25

c. 1

d. 0.5

20. A line M through A is drawn parallel to BD. Point S

moves such that its distances from the line BD and the

vertex A are equal. If locus of S cuts M at and and

AC at then area of is

a. sq. unit

b. sq. unit

c. 1 sq. unit

d. 2 sq. unit

2

2

1 1log

2 1

+ ++ − +

x x

x x

e eC

e e

2

2

1 1log

2 1

− ++ + +

x x

x x

e eC

e e

4 2

4 2

1 1log

2 1

+ ++ − +

x x

x x

e eC

e e

2

0

1 4sin 4sin2 2

x xdx

π

+ −∫4π −

24 4 3

3

π− −

4 3 4−

4 3 43

π− −

( ) 1( ) 200.

2

dp tp t

dt= − ( )0 100,p =

/ 2400 300

te−

/ 2300 200

te−−/ 2

600 500t

e−/ 2400 300 te−−

60° 3 1.x y+ =

3 2 3 3 0y x+ + − =

3 2 3 3 0y x− + + =

3 3 2 3 0y x− + + =

3 3 2 3 0y x+ − + =

1

1 2 1: ,

3 1 2

x y zL

+ + += =

2

2 2 3:

1 2 3

x y zL

− + −= =

1L 2L

ˆˆ ˆ7 7

99

i j k− + +

ˆˆ ˆ7 5

5 3

i j k− − +

ˆˆ ˆ7 5

5 3

i j k− + +

ˆˆ ˆ7 7

99

i j k− −

1L 2L

2

75

7

75

13

75

23

75

2C

1C

1C 2 ,C

2 2 2 2

2 2 2 2

PA PB PC PD

QA QB QC QD

+ + ++ + +

2T 3T

1,T 1 2 3T T T∆

1

2

2

3

203Mock Test-5

ANSWER & SOLUTIONS

JEE-Main

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

d a c c b b c a a b

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

c b d c b c b d a d

21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

c c a b d b a b a c

1. (d)

⇒ for domain

2. (a)

⇒

⇒

⇒

3. (c)

4. (c) It is skew-symmetric.

5. (b) The sequence is a G.P. with common ratio

Now from

6. (b) We have

Therefore coefficient of in the given expression

Coefficient of in

7. (c)

Here

.

Trick: The sum of this series upto 4 terms is 1.359…and

this is value of e/2 approximately.

8. (a) Four first prizes can be given in ways since first

prize of Mathematics can be given in 20 ways, first prize

of Physics also in 20 ways, similarly first prizes of

Chemistry and English can be given in 20 ways each.

(Note that a boy can stand first in all the four subjects).

Then two second prizes can be given in ways since a

boy cannot get both the first and second prizes. Hence the

required number of ways

9. (a)

Therefore, for the number ends at unit place 7,

9, 3, 1, 7, …..

will be divisible by 5 if it end at 5 or 0.

But it cannot end at 5. Also for end at 0. For this m and n

should be as follows

m n

1

2

3

4

For any given value of m, there will be 25 values of n.

Hence, the probability of the required event is

Note: Power of prime numbers have cyclic numbers in

their unit place.

10. (b)

.

11. (c)

2 2 2y x= −

2log 2 2xy = − 2 2 1x x− ⇒ <

4 2log log ( 8 ) 0x x+ − =

( )0

24 log 8x x= + −

⇒ 12 8x x= + −

⇒ 24 8 2 8x x x x= + + − +

22 8 2 4x x x+ = +

⇒ 2 28 4 4x x x x+ = + +4 4x =

⇒ 1.x =

cosh( ) cosh( )i iα β α β+ − −

cosh cosh ( ) sinh sinh( )i iα β α β= +

cosh cosh ( ) sinh sinh ( )i iα β α β− +

2sinh sinh iα β= 2 sinh sin .i α β=

1.

3

20(1 ) 2[1 (1/3) ]

,1 1 (1/3)

na r

r

− −− −

20

13 1 .

3

= −

1 2( 3) ( 3) ( 2)n nx x x

− −+ + + + +3 2 1( 3) ( 2) .... ( 2)n n

x x x− −+ + + + +

( 3) ( 2)( 3) ( 2)

( 3) ( 2)

n nn nx x

x xx x

+ − += = + − +

+ − +

1 2 1 3 2 1( .... )n n

n n n nx ax x a x a a

x a

− − − −−= + + + +

−Q

rx

= rx [( 3) ( 2) ]n n

x x+ − +

3 2 (3 2 )n n r n n r n n r n r

r r rC C C− − − −= − = −

1 2 3 4

1! 3! 5 ! 7 ! (2 1) !

nS

n= + + + + + +

−K K

1 2 1 (2 1) 1.

2 (2 1) ! 2 (2 1) !n

n nT

n n

− += =

− −

1 1 1

2 (2 2) ! (2 1) !n n

= +

− −

⇒1 1

1

2 2 2 2n

e e e e eS T

− − + −= = + =

∑

420

219

4 220 19 .= ×

1 2 3 47 7,7 49,7 34,7 2401,.....= = = =

7 ,r r N∈

∴ 7 7m n+

4r 4 2r −

4 1r − 4 3r −

4 2r − 4r

4 3r − 4 1r −

100 25 1

100 100 4

×=

×

2sin( ) sin( )cosecπ θ π θ θ+ −

2

1sin sin 1

sinθ θ

θ= − = −

2 1sinh (cosh 2 1)

2x x= −

Mathematics204

12. (b)

From figure, we can deduce tan cotH l α β= .

13. (d) 0;

Hence no value of k can make

14. (d) (1 )xy x= +

Taking log on both sides, log log(1 )y x x= +

Differentiating w.r.t. x, we get

1 1

log(1 )(1 )

dyx x

y dx x= + +

+

Thus (1 ) log(1 )1

xdy xx x

dx x

= + + + +

15. (b) If is maxima, then its reciprocal

will be minima.

Let

∴

On putting

⇒

⇒

⇒

which is positive.

At is minimum.

So will be maximum.

16. (c) Put then

17. (b)

Given or . . .(i)

There are two points on the curve (a, a),(– a,– a)

The equation of the line at (a, a) is,

therefore, equation of the tangent at

is The interception of line

with x-axis is 2a and with y-axis is 2a.

Required area

18. (d)

⇒

⇒

⇒

19. (a) Distance of point (–2, 3) from the line 5x y− = is

2 3 5 105 2.

2 2

− − − −= =

20. (d) Obviously the centre of the circle is (4, 2) which

should be the middle point of the ends of diameter.

Hence the other end is (11, 2).

21. (c)

2

0(0 ) lim (2 ) 0

xf k x x

→ −− = − =

0(0 ) lim cos 1

xf x

→ ++ = =

∴ (0) cos 1f x= =

(0 ) 1.f − =

1 tan

x

x x+

1 tanx x

x

+

1 tan 1tan

x xy x

x x

+= = +

2

2

1sec ,

dyx

dx x= − +

2

2 3

22sec sec tan

d yx x x

dx x= +

0,dy

dx= 2

2

1sec 0x

x− + =

2

2

1sec x

x=

2 2cosx x=

cosx x=

∴2

2

2 3

22sec tan

cos

d yx x

dx x= +

22sec (sec tan ),x x x= +

cos ,x x=1 tanx x

x

+

1 tan

x

x x+

11 log ,t x dt dx

x= + ⇒ =

1/ 2 1/ 2

1/ 22 2(1 log ) .

1 log

dx dtt c x c

tx x= = + = + +

+∫ ∫

2xy a=2a

yx

=

( , )

( )a a

dyy a x a

dx

− = −

( , )

2

2( )

a a

ax a

x

− = −

( )y a x a− = − −

( , )a a 2 .x y a+ = 2x y a+ =

∴ 212 2 2 .

2a a a= × × =

logdy

x xdx

=

logdy x xdx=

logdy x xdx=∫ ∫2 2

log .2 4

x xy x c= − +

x O

y

1

3

(x,y)Q

P(h,k)

y2 = 4x

(0,0)

2a

2a

X

(–a,–a)

Y α β

A O

l

P

T

H

205Mock Test-5

By section formula,

Substituting in

Or is required locus.

22. (c) Squaring

we get

⇒

⇒

23. (a) Here, three given lines are coplanar if they have

common perpendicular

Let d.c.'s of common perpendicular be l, m, n

⇒ . . .(i)

. . .(ii)

and . . .(iii)

Solving (ii) and (iii), we get

⇒

Substituting in (i), we get

⇒

24. (b) .

25. (d) Under condition,

26. (b) Since a, b, c and u, v, w are the vertices of two

triangles.

Also,

and . . .(i)

Consider

Applying

[from Eq. (i)]= 0

Hence, two triangle are similar.

27. (a) Given ,

Since every minor of order 3 in A is 0 and there exists a

minor order 3 i.e. in A which is non-zero. Thus,

rank = 2.

28. (b) We have,

Coefficient of is

29. (a) Total number of ways to arrange 3 boys and 2 girls are

5!. According to given condition, following cases may

arise.

B G G B B

G G B B B

G B G B B

G B B G B

B G B G B

So, number of favourable ways 5 3! 2! 60= ¥ ¥ =

Required probability

30. (c) We know that

then the Imaginary part of be

,

0 0,

4 4

x yh k

+ += =

∴ 4 , 4x h y k= =2 2 24 , (4 ) 4(4 )y x k h k h= = ⇒ =

2y x=

( ) 0,a b c+ + =rr r

2 2 2 2 2 2 0a b c a b b c c a+ + + ⋅ + ⋅ + ⋅ =r r rr r r r r r

2 2 2| | | | | | 2( ) 0a b c a b b c c a+ + + ⋅ + ⋅ + ⋅ =

r r rr r r r r r

2( ) 3a b b c c a⋅ + ⋅ + ⋅ = −r rr r r r

⇒3

2a b b c c a⋅ + ⋅ + ⋅ = −

r rr r r r

1 1 1 0ll mm nn+ + =

2 2 2 0ll mm nn+ + =

3 3 3 0ll mm nn+ + =

2 3 2 3 2 3 3 2 2 3 3 2

l m nk

m n n m n l n l l m l m= = =

− − −

2 3 2 3 2 3 3 2 2 3 3 2( ), ( ), ( )l k m n n m m k n l n l n k l m l m= − = − = −

1 2 3 2 3 1 2 3 3 2 1 2 3 3 2( ) ( ) ( ) 0l m n n m m n l n l n l m l m− + − + − =

1 1 1

2 2 2

3 3 3

0

l m n

l m n

l m n

=

⇒

1 1 1

2 2 2

3 3 3

0

l n m

l n m

l n m

− =

qpqpqp ∧≡∧≡∨ )(~)(~~~))(~(~

23

λ− =

⇒2

3λ = −

( )1c r a rb= − +

( )1w r u rv= − +

111

a ub vc w

3 3 1 2(1 ) R R r R rR→ − − +

( ) ( ) ( )

11

1 1 1 1

a ub v

c r a rb w r u rv r r=

− − − − − − − − −

11

0 0 0

a ub v=

2 3 1 4

0 1 2 1

0 2 4 2

A

= − − −

2 2 3( 2 )R R R→ +

2 3 1 4

0 0 0 0

0 2 4 2

A

= − −

2 3

0 2

−

2 1(1 ...) [(1 ) ] (1 )n n nx x x x− − −+ + + = − = −2

0 1 2 ... ( 1) .n n n n n n

nC C x C x C x= − + + + −n

x ( 1) ( 1) .n n n

nC− = −

∴60 1

120 2= =

1 1 sintan (cos sin ) log ,

4 4 1 sin

ii

π θθ θ

θ− + + = + −

(cos ) 0θ >1tan (cos sin )iθ θ− +

11tanh (sin )

2θ− 1 1 1

tanh log2 1

xx

x

− + = − Q

Mathematics206

JEE Advance Paper-I

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

2560 3 1004 3954 139 4 112 2880 a d

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

d bcd b ad abc b bd c b a

1. (2560) Let the equation of the variable plane be

. . .(i)

Given that the plane is at a distance p form (0, 0, 0)

or . . .(ii)

Also, the plane (i) meets the axes in A, B and C. So the

coordinates of O, A, B and C are (0, 0, 0), (a, 0, 0), (0, b,

0) and (0, 0, c) respectively

Let be the centroid of the tetrahedron OABC,

then

Similarly and

or

Substituting these values in equation (ii), we get

Or

2. (3) Comparing the given parabola (i.e., 8x) with

Since, normal at to the parabola is

Here, and

Equation of normal is

. . . (i)

Solve equation (i) and then

and

Then and

Hence, point of intersection of normal and parabola are

and therefore normal meets the parabola

at and length of normal chord is distance

between their points

(given)

3. (1004) Given that

Thus,

and . . .(i)

. . .(ii)

But

from equation (i)

. . .(iii)

Putting the value of from equation (iii) in (ii), then

(given)

4. (3954) Using the formula for the sum of a geometric

progression, we find

Hence, the coefficient of

and

1x y z

a b c

+ + =

∴2 2 2

1

1 1 1

p

a b c

= + +

2 2 2 2

1 1 1 1

p a b c= + +

( , , )x y z

1 1(0 0)

4 4x a a= + + =

1

4y b=

1

4z c=

4 , 4 , 4a x b y c z= = =

2 2 2 2

1 1 1 1

16 16 16p x y z= + +

2 2 2 216x y z p− − − −+ + =

∴ 16λ =⇒ 160 160 16 2560λ = × =

2y x=2 4y ax=

∴ 4 8a =

∴ 2a =

1 1( , )x y 2 4y ax=

11 1( )

2

yy y x x

a− = −

12x =

14y =

∴4

4 ( 2)4

y x− = − −

⇒ 4 2y x− = − +

⇒ 6 0x y+ − =2 8y x= 2 8(6 )y y= −

⇒ 2 8 48 0y y+ − =

⇒ ( 12)( 4) 0y y+ − =

∴ 12y = − 4y =

18x = 2x =

(18, 12)− (2, 4)

(18, 12)−

2 2(18, 12) ( 12 4) 16 2PQ= = − + − − =

λ=

∴ 2 5 / 2λ = ( )2.5 3 .≈

( ) , ( / ) ( / ) 1P A P B A P B Aα α′ ′= = = −

( ) 1 ( )P A P A′ = − 1 α= −

( / ) 1 ( / ) 1 (1 )P B A P B A α α′ ′ ′= − = − − =

∴( ) ( ) ( )

( / )( ) ( )

P A B P B P A BP A B

P B P B

′∩ − ∩′ = =

( ) ( ) ( / )

( )

P B P A P B A

P B

−=

( )( / )

( )

P A BP B A

P A

∩=

Q

( ) (1 )

( )

P B

P B

α α− −=

( ) ( ) ( / ) ( ) ( / )P B P A P B A P A P B A′ ′= ⋅ + ⋅

(1 ) (1 )α α α α⋅ − + − ⋅

2 (1 )α α= −

( )P B

2 (1 ) (1 ) (1 ) 1( / )

2 (1 ) 2 (1 ) 2P A B

α α α α α αλ

α α α α− − − −′ = = = =

− −

∴1

2008 2008 10042

λ = × =

1000 999 2 998 1000(1 ) (1 ) (1 )x x x x x x+ + + + + + +K

10011001

1000 1000

1001 1001

(1 ) 1 (1 )1 (1 )

(1 )1

1(1 )1

x xx xx x

x xx xx

xx

+ − + − + + = = = + −+ − − ++

50 1001

50

1001!

50!951!x C= =

∴ 1001, 50λ µ= = v 951=

∴ 2 3 1001 100 2853 3954vλ µ+ + = + + =

207Mock Test-5

5. (139) ,

,

And

Hence, required value is

6. (4)

and

. . .(i)

. . .(ii)

. . .(iii)

⇒

7. (112)

8. (2880)

9. (a) Set A has 4 elements

Set B has 2 elements

Number of elements in set

Total number of subsets of

Number of subsets having 0 elements

Number of subsets having 1 element each

∴ Number of subsets having 2 elements each

Number of subsets having at least 3 elements

10. (d) with a, b of same sign.

If or

⇒

⇒

Hence real or purely imaginary number cannot satisfy

11. (d)

(For purely imaginary)

,

12. (b,c,d)

Null matrix if n is a multiple of 3

13. (b) Given m is A. M. between and n

Given in G.P.

Q 1 7 7sin sin

6 6 6

π π ππ− = − = −

1 2 2cos cos

3 3

π π− =

1 5 5tan tan

4 4 4

π π ππ− = − =

1 3cot cot

4 4 4

π π ππ− − = − =

1 2 3216 27 28 200

6 3 4 4

π π π ππ ×− + × + × + ×

36 18 7 150 139= − + + + =

| | | | | | 1= = =rr r

a b c

× + × = + +r r rr r r r

a b b c pa qb rc

( ) ( ) ( )⋅ × = + ⋅ + ⋅r rr r r r r

a b c p q a b r a c1

[ ]2

=rr r

a b c

[ ]2 2

+ + =rr rq r

p a b c

02 2+ + =

p rq

[ ]2 2+ + =

rr rp qr a b c

= = −p r q

⇒2 2 2

2

24

+ +=

p q r

q

∴2 2

2 22 , 28 8

y yf x x xy

+ − =

2 22 2

2 22 28 8

y yx x

= + − −

∴ (60, 48) (80, 48) (13, 5)f f f+ +

2 2 2 2 2 2(60) (48) (80) (48) (13) (5) 36 64 12= − + − + − = + +

112=

/ (cos ( sin ) cos )tan

/ ( sin cos sin )

dy dy dt a t t t tt

dx dx dt a t t t t

− − −= = =

− + +

⇒2

2(tan ) (tan )

d y d d dtt t

dx dx dt dx= = ⋅

32 1 sec

seccos

tt

at t at= =

2

2

/3

8 24

/ 3t

d y

dx a aπ

π π=

= =

∴2

2

/3

120 120 24 2880

t

d ya

dxπ

π=

= × =

∴ ( ) 4 2 8× = × =A B

∴ 8( ) 2 256× = =A B

08 1= =C

18 8= =C

2

8! 8 78 28

2!6! 2

×= = = =C

256 1 8 28 256 37 219= − − − = − =

2( )P x ax b= +2 2( ( )) ( )P P x a x b b= + +

x R∈ ix R∈2x R∈

( )P x R∈

⇒ ( ( )) 0P P x ≠

( ( )) 0.P P x =

2 3 sin 1 2 sin

1 2 sin 1 2 sin

θ θθ θ

+ +×

− +

i i

i i

22 6sin 0θ− =

2 1sin

3θ =

1sin

3θ = 1 1

sin3

θ −=

2 3 4 2

3 4 5 3

2 3 2 4

ω ω ω ωω ω ω ω

ω ω ω

+

+

+ + +

=

K

K

M M M M M

K K

n

n

n n n

P

4 6 5 7 9

5 7 9

2

4 6 2 4 2 6

ω ω ω ω ωω ω ω

ω ω ω ω+ + + +

+ + +

+ + =

+ +

K K K

K K K K

M M M K

K K K Kn n n n

P

2 =P

l

⇒ 2m n= +l

1 2 3, , , ,G G G nl

1/ 4

4n nr r

= ⇒ = l l

Mathematics208

So,

14. (a,d)

15. (a,b,c)

Rearranging, we get

Rearranging again,

16. (b) Case-1 Any 5 – digit number > 6000 is all 5-digits

number

Total number > 6000 using 5-digits = 5! = 120

Case-2: Using 4-digits

Can be 6.7 or 8 4 ways 3 ways 2 ways

i.e., 3 ways Total number

Total ways

17. (b,d)(a)

(b)

(c)

(d)

18. (c)

⇒

19. (A)

Using

(B) For domain

Case (i):

∴ 2 3

1 2 3, ,G G r G r= = =l l l

4 4 4 4 4 4 8

1 2 32 [1 2 ]G G G r r r+ + = + +l

2

4 1 2n n n = ⋅ + +

ll l l

2 23 3 2 2

2

( )1 (2 ) 4

n nn n n m m n

+ = + = = =

ll l l l

l l

( 1) ( 1)42 2 2 2 22

1 0

( 1) ((4 4) (4 3) (4 2) (4 1) )k k nn

n

k r

S k r r r r+ −

= =

= − + + + − + − +∑ ∑( 1)

0

(2(8 6) 2(8 4))n

r

r r−

=

= + + +∑( 1)

0

(32 20)n

r

r−

=

= +∑

16( 1) 20n n n= − +

4 (4 1)n n= +1056 for n 8

1332 for n 9

==

=

2 2log 3 ( 1) log 4 2( 1)= − = −x x x

⇒2log 3 2 2= −x x

⇒2

2

2 log 3=

−x

3

3

3

2 log 22

1 2 log 2 12

log 2

−= =

−−x

3 4

3 4

4

1

log 4 log 3 1.

1log 4 1 1 log 31

log 3

= = =− −−

x

3 4 3 2 72= × × × =

120 72 192= + =

1

1 1 112 4 4

1 1 3 1 1 1 1 1 3 1 1 1 8

2 4 4 4 4 2 2 4 4 2 4 4

cXP

X

× × =

× × + × × + × × + × ×

8 1 1 1

Exactly twoengingesarefunctioning 732 4 4 28X 8

32

P

− × × = =

2

1 1 3 1 1 1 1

54 2 4 2 4 2 4

1 1 1 1 3 1 1 1 3 8

4 2 4 2 4 2 4 2 4

XP

X

× + × + × = = × + × + × + ×

1

1 1 1 1 3 1 3

72 4 4 4 4 4 4

1 3 3 1 1 1 3 1 3 16

2 4 4 4 4 4 4 4 4

XP

X

× + × + × = = × + × + × + ×

2sin 2sin cos

2sin 2sin cos

P P P

P P P

−

+

⇒1 cos 3

1 cos 32

P

P

−=

+

29cos

35P =

1 5 7 8 6

2 2 2 35∆ = × × ×

⇒ 6∆ = ⇒

23

4

∆

2 2 2

2i(x iy) 2i(x iy)z

1 (x iy) 1 (x y 2ixy)

+ += =

− + − − +2 21 x y− =

2

2ix 2y 1Z .

y2y 2ixy

−= = −

−

Q1 1

1 y 1 1or 1.y y

− ≤ ≤ ⇒ − ≤ − − ≥

x 2

2(x 1)

8.31 1

1 3

−

−− ≤ ≤

−

⇒x x 2

2x 2

3 31 1.

1 3

−

−

−− ≤ ≤

−x x 2

2x 2

3 31 0

1 3

−

−

−− ≤

−

P

Q R

5/2 7/2

1B 3A

2B 2A

3B 1A

1/ 2 1/ 2= − =x x

O

/ 6π

209Mock Test-5

Case (ii):

So,

(C)

(D)

Increasing, when

20. (A)

So, one solution.

(B) Let (a, b, c) is direction ratio of the intersected line, then

We must have

(C) Let

⇒ k can take value 2, 3, 4, 5.

(D)

⇒


1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

1 36 64 5522 5 1352 4 16 a a

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

c a c d a b b c a c

1. (1) Let

Let

2. (36) Let

Put

Then,

We, get Then,

3. (64) We have,

⇒

Or

Putting we have,

∴

⇒ i.e.,

⇒x x 2

2x 2

(3 1)(3 1)0

(3 1)

−

−

− −≥

−

⇒ x ( ,0] (1, ).∈ −∞ ∪ ∞

x x 2

2x

3 31 0

1 3 2

−−+ ≥

− −

⇒x 2 x

x x 2

(3 1)(3 1)0

(3 .3 1)

−

−

− +≥

−

⇒ x ( , 1) [2, ).∈ −∞ ∪ ∞

x ( , 0] [2, ).∈ −∞ ∪ ∞

1 1 3R R R→ +

0 0 2

f ( ) tan 1 tan

1 tan 1

θ θ θθ

= −

− −

2 22(tan 1) 2 sec .θ θ= + =

1/ 2 3 / 2 1/ 23 15f (x) (x) (3x 10) (x) 3 (x) (x 2)

2 2′ = − + × = −

x 2.≥

'( ) 0, (0, / 2)π> ∀ ∈f x x

(0) 0and ( / 2) 0π< >f f

4 0+ + =ak b c

4 2 0+ + =a kb c

28 4 2 16= =

− − −a b c

k k k

22(8 ) 2(4 2 ) ( 16) 0− + − + − =k k k

⇒ 2,4.=k

( ) | 2 | | 1| | 1| | 2 |= + + + + − + −f x x x x x

1=

+∫ ∫dy

dxy

( ) 2 1= −xf x e

⇒ (ln 2) 3=f

10( ) logf x x=

⇒10

1( ) logf x e

x′ = ⋅

4, 0.04x xδ= =

∴ ( ) ( ) ( )f x x f x x f xδ δ ′+ = + ⋅

⇒10 10 10

1log ( ) log (0.04) logx x x e

xδ+ = + ×

⇒10 10 10

1log 4.04 log 4 (0.04) log e

x= + × ×

0.6021 (0.01)(0.4343)= +

0.6021 0.004343 0.606443= + = 1≈

( tan cot )I x x dx= +∫(sin cos )

(sin cos )

x xdx

x x

+= ∫

sin cosx x t− =

⇒ 21 sin 2x t− =

∴ (cos sin )x x dx dt+ =

1 1

222 sin 2 tan

11

2

dt tI t c

tt

− − = = = +

− −

∫

1 sin cos2 tan

sin 2

x xc

x

− − = +

1 tan 1

2 tan2 tan

xc

x

− − = +

2, 2a b= =

⇒4 5 4 32 36a b+ = + =

3 3

2 2

/8

/

d y dx

d y dx=

2

2ln 8

d yx c

dx= +

2ln 8y x c= +

0x =

2log (0) ln1 0c y= = =

2ln 8y x=

8

2

xy e=

8

18

xe

y D= +

4x− 4x

2 4x −6

4 2x−

2− 1− 1 2

Mathematics210

Again putting then

⇒

⇒

Putting we have

Hence,

4. (5522) Since and are the roots of the equation

Then

and coordinates of the vertices of the given triangle are

(0, 0) and

Hence, the required area of triangle

On comparing,

or

⇒

5. (5) Equation of chord of contact is

and

Then

Area of

For maximum or minimum then we get

( as P is outside the circle)

And

Area of triangle is also maximum at

6. (1352) Let PV of P, A, B and C are and

respectively and be the circumcircle of the

equilateral triangle ABC.

Then, . . .(i)

and unit . . .(ii)

Now,

Similarly and

[From equation (i) and (ii)]

0,x =1

1(0)

8y D= +

10

8D= +

∴ 1

8D = −

8

1

1

8 8

xe

y = − ⇒8

64 8

xe x

y E= − +

0,x = 1 1(0) 0

64 8y E= − + =

∴ 1 1 7

8 64 64E = − =

8 87 ( 8 7)

64 8 64 64

x xe x e x

y− +

= − + = ⇒ 64λ =

1m

2m

2 ( 3 2) ( 3 1) 0x x+ + + − =

1 2 1 2( 3 2), ( 3 1)m m m m+ = − + = −

∴ 2

1 2 1 2 1 2( ) 4 (3 4 4 3 4 3 4) 11m m m m m m− = + − = + + − + =

1( / , )c m c

2( / , )c m c

1 2

1

2

c cc c

m m= × − ×

2 12 2

1 2 1 2

1 1 1 1

2 2

m mc c

m m m m

− = − =

21 11

2 ( 3 1)c=

−

2 21 11( 3 1) 33 11

2 4( 3 1)( 3 1)c c

+ += = − +

33 11,

4 4a b= =

11 33,

4 4a b= =

∴ 2 2 33 11 44 11

16 16 16 4a b+ = + = =

2 2 112008( ) 2008

4a b+ = × 502 11 5522= × =

( )QR 26 8 0x y r+ − =

2 2

2 2

6 6 8 8 100

10(6 8 )

r rPM

⋅ + ⋅ − −= =

+

2 2

2 2

0 0

10(6 8 )

r rOM

+ −= =

+

42 2 22 2 ( ) ( ) 2

100

rQR QM OQ OM r

= ⋅ = − = −

1

2QPR QR PM∆ = ⋅ ⋅

∴24

21001

(say) 22 100 10

rrr

− ∆ = ⋅ − ⋅

∴2 2 3

2 (100 )(say)

1000

r rz

−∆ = =

∴ 2 2 2 2 31 3(100 ) ( 2 ) (100 ) 2

1000

dzr r r r r

dr= ⋅ − ⋅ − + − ⋅

2 22 22 (100 )

100 3 1000

r rr r

−= − −

0,dz

dr=

5,r = 10r ≠2

2

5r

d zve

dr=

= −

∴ 5.r =

,, a bprrr

cr

(0)Or

03

a b c+ +=

rr r

26| | | | | | | |

3p a b c= = = =

rr r r

2 2 2 2| | | | 2PA a p a p a p= − = + − ⋅uuur r r r r

2 2 2| | 2PB b p b p= + − ⋅uuur r r

2 2 2| | 2PC c p c p= + − ⋅uuur r r

∴ 2 2 2 2 2| | 3 2 ( )PA a b c p p a b cΣ = + + + − + +uuur rr r r

26 0p= −

22(26)

6 2(26) 13523

= = =

(6, 8)

Q

R

O

r r

M

P

(0, 0)

211Mock Test-5

7. (4) Any point on the first line in symmetrical form is

If the lines are coplanar, this

point must lie on both the planes which determine the

second line.

. . .(i)

and . . .(ii)

From equation (i), we get

Now substituting equation (ii), then

8. (16) Let the equation of circle

. . .(i)

and the equation of the rectangular hyperbola is

. . .(ii)

Put and in equation (i)

Then

This equation being fourth degree in t. Let the roots be

then

9. (a)

10. (a)

Similarly,

11. (c) From graph,

which holds if

12. (a)

..........

Summing up all terms

Hence,

13. (c)

where

14. (d)

(3 4, 5 6, 2 1).r r r− − − +

⇒ 3(3 4) 2(5 6) 2 1 5 0r r r− − − − + + =

2(3 4) 3(5 6) 4( 2 1) 0r r r k− + − + − + − =

2r =

2r = 4k =

2 2 2 2 0x y gx fy k+ + + + =

2xy c=

x ct=c

yt

=

22 2

2

22 0

c fcc t gct k

t t+ + + + =

⇒ 2 4 3 2 22 2 0c t gct kt fct c+ + + + =

1 2 3 4, , ,t t t t 1 2 3 4 1t t t t =

∴ 1 2 3 416 16 1 16t t t t = × =

21 1

2

2

1tan ( ) tan.2

11

xx

xyx x

x

− −

+ −= −

−

⇒3

1 1

2

3tan ( ) tan

1 3

x xy

x

− − −=

− ⇒

3

2

3

1 3

x xy

x

−=

−

1

3

h

AQ=

⇒ 3AQ h=

BQ h=

⇒3

hCQ =

∴ ( 3 1) 3

1

3

AB AQ BQ h

BC BQ CQ hh

− −= = =

− −

1p = −

⇒1

lim ( ) 1x

g x+→

= − ⇒0

lim (1 ) 1h

g h→

+ = −

⇒0

lim 1log cos

n

mh

h

h→

= −

⇒1 1

0 0lim lim 1,

( tan ) tan

n n

h h

n h n h

m h m h

− −

→ →

⋅ = − = − ⋅ −

2.n m= =

( 1) log( ( 1) log log( ( )) log ( )+ = + = + = +g x f x x f x x g x

⇒ ( 1) ( ) log+ − =g x g x x

⇒2

1( 1) ( )′′ ′′+ − = −g x g x

x

1 11 4

2 2

′′ ′′+ − = −

g g

1 1 42 1

2 2 9

′′ ′′+ − + = −

g g

2

1 1 4

2 2 (2 1)

′′ ′′+ − − = − − g N g N

N

2

1 1 1 14 1 ... .

2 2 9 (2 1)

′′ ′′+ − = − + + + − g N g

N

2 2

4 2 4 2

( 1) ( 1)

1 1

− −− = =

+ + + +∫ ∫x x

x x

e e ZJ I dx dz

e e z z

= xz e

2

2

11

1 1ln

2 111

−

−

− + − = = + + + + −

∫x x

x x

dze ez

ce e

zz

∴2

2

1 1ln .

2 1

− +− = + + +

x x

x x

e eJ I c

e e

2

0

1 4sin 4sin2 2

x xI

π

= + −∫0

1 2sin2

xI dx

π

= −∫

3

0

3

1 2sin 1 2sin2 2

xI dx dx

ππ

π

π = − + − − ∫ ∫

3

03

1 4cos 4cos2 2

x xx x

ππ

π

= + + − −

0 1

x–1 –x+1

A QCB

h

P

30° 45° 60°

Mathematics212

15. (a) Rearranging the equation we get,

. . .(i)

Integrating (1) on both sides we get

where k is a constant of integration.

Using we get

the relation is

16. (b) Inclination of line is

Inclination of line

Slope of line

Equation of = Line L

⇒

17. (b)

Hence unit vector will be

18. (c) Plane is given by

distance

19. (a) Let A, B, C and D be the complex numbers,

and respectively.

20. (c)

[as A is the foucs,

is the vertex and BD is the directrix of parabola].

Also is latus return

∴ Area of

3 31 4 1 4 0

3 2 3 2

π ππ

= + − − − − −

22 3 4 2 3 4 3 4

3 3 3I

π π π= + − − + = − −

( ) 1

( ) 400 2

dp tdt

p t=

−

/ 2( ) 400 ,

tp t k e= +

(0) 100,p = 300k = −

∴ 1/ 2( ) 400 300P t e= −

3 1x y+ = 150°

150 60L = ° ± ° 210 , 90= ° °

1tan 210 tan 30

3L = ° = ° =

12 ( 3)

3y x+ = −

⇒ 3 3 2 3 0y x− + + =

ˆˆ ˆ3 1 2 7 5

1 2 3

i j k

i j k= − − +

ˆˆ ˆ7 5.

5 3

i j k− − +

( 1) 7( 2) 5( 1) 0x y z− + − + + + =

⇒ 7 5 10 0x y z+ − + =

⇒ 1 7 5 10 13.

75 75

+ − += =

2, 2, 2i−

2i−

⇒2 2 2 2

2 2 2 2

PA PB PC PD

QA QB QC QD

+ + ++ + +

2 2 2 2 2

1 1 1 1 1

22 2 2 212 2 2 2

| 2 | | 2 | | 2 | | 2 | | | 2 3

| | 2 4| 2 | | 2 | | 2 | | 2 |

z z z i z i z

zz z z i z i

− + − + − + − += = =

++ + − + − + −

2AG =

∴ 1 1

1

2AT TG= =

1T

2 3T T

∴ 2 3

14

2T T = ×

1 2 3

1 1 41

2 2 2T T T∆ = × × =

D

G

C

B A

M

T1 T2

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