Jee Adv 2015 Paper 1

Sri Chaitanya IIT Academy., India

#Corporate Office: Plot No: 304, Kasetty Heights, Ayyappa Society, Madhapur Hyd

Email: [email protected], Web: www.srichaitanya.net

2015 JEE-ADVANCED PAPER-1_QPAPER WITH SOLUTIONS

INSTRUCTIONS General 1. This sealed booklet is your Question Paper. Do not break the seal till you are instructed to do so. 2. The question paper CODE is printed on the left hand top corner of this sheet and right hand top corner of the back cover of this booklet. 3. Use the Optical Response Sheet (ORS) provided separately for answering the questions. 4. The ORS CODE is printed on its left part as well as the right part. Ensure that both these codes are identical and same as that on the question paper booklet. If not, contact the invigilator. 5. Blank spaces are provided within this booklet for rough work. 6. Write your name and roll number in the space provided on the back cover of this booklet. 7. After breaking the seal of the booklet, verify that the booklet contains 32 pages and that all the 60 questions along with the options are legible. QUESTION PAPER FORMAT AND MARKING SCHEME: 8. The question paper has three parts: Physics, Chemistry and Mathematics. Each part has three sections. 9. Carefully read the instructions given at the beginning of each section. 10. Section 1 contains 8 questions. The answer to each question is single digit integer ranging from 0 to 9 (both inclusive).

Marking Scheme: +4 for corrected answer and 0 in all other cases. 11. Section 2 contains 10 multiple choice questions with one or more than one corrected option. Marking Scheme: +4 for correct answer, 0 if not attempted and -2 in all other cases. 12. 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, +2 for correct answer, 0 if not attempted and -1 in all other cases.

OPTICAL RESPONSE SHEET: 13. The ORS consists of an original (top sheet) and its carbon-less copy (bottom sheet). 14. Darken the appropriate bubbles on the original by applying sufficient pressure. This will leave an impression at the corresponding place on the carbon-less copy. 15. The original is machine-gradable and will be collected by the invigilator at the end of the examination. 16. You will be allowed to take away the carbon-less copy at the end of the examination. 17. Do not tamper with or mutilate the ORS. 18. Write your name, roll number and the name of the examination center and sign with pen in the space provided for this purpose on the original. Do not write any of these details anywhere else. Darken the appropriate bubble under digit of your roll number.




PART-1: PHYSICS (SINGLE DIGIT INTEGER TYPE QUESTIONS)

This section contains 8 Single digit integer type questions ranging from 0 to 9, both inclusive.(Each question correct marking (+4), No Negative (0))

1. A youngs 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 2x p m d , where 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

1S

2S

x

d

d Air

Water

Key: 3

Sol: 2 2 2 2 2x P m d

Optical path differences,

2 2 2 21 2x S P S P x d x d

2 2m x d 1

2 2 2 2 2m x d 1 2

2 2 16 8x d 19 3

2 2 8x d 19

2 2 2 2 2 2 2 21m x d 9m x d9

P 3

S1

S2

d x1

4 / 3 d




air

3015

20I1

I2O

air

20

30 20I1

O

40

1.5medium(7/6)

medium(7/6)

2. Consider a concave mirror and a convex lens (refractive index= 1.5) of focal length

10cm each, separated by a distance of 50cm 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 1M . When the set-up is kept in a

medium of refractive index 7/6, the magnification becomes M2. The magnitude 21

MM

is

50 cm

15 cm

Key: 7 Sol: 1st case: Reflection from mirror 1 1 1

f v u

1 1 1 v 3010 v 75

For lens 1 1 1

f v u

1 1 110 v 20

v 20 1M 2 1 2 (in air) 2nd case : For mirror v 30 For lens

1 1 3 / 2 21v 20 7 / 6 R

1 1 2 2 4v 20 7 10 70

1 4 1 v 140v 70 20

, 2M 2 140 / 20 14 , 2

1

M 14 7M 2




3. An infinitely long uniform line charge distribution of charge per unit length lies

parallel to the y-axis in the y-z plane at 32

z a (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 centre at the origin is 0

Lz an

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

of n is

32

a

L

A B

x

CD

yO

z

Key: 6 Sol: Given rectangular surface ABCD subscribes an angle 060 at the infinity long uniform

line of change

0 0

q L6 6

n 6 4. Consider a hydrogen atom with its electron in the thn orbital. An electromagnetic

radiation of wavelength 90nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4eV , then the value of n is ( 1242 )hc eVnm

Key: 2

Sol: Energy incident incinc

hCE

1242 13.8eV90

By law of conservation of energy,

Energy of the ejected electron 213.6eV

n Energy incident

213.6eV10.4eV 3.8eV

n

2213.6eV 3.4 n 4

n n 2




5. 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 planets

gravity is 1/4th of its value at the surface of the planet, if the escape velocity from the

planet is esev v N , then the value of N is (ignore energy loss due to atmosphere)

Key: 2

Sol: 22 hgR g R h GM

22 ggR R h h R4

By law of conservation of energy

21 GMm GMmmv 02 R R h

2 2

2e ee

mv mv1 1 1mv2 N 2 2 2

2

2ee

mv1 1 1mv2 N 2 2

N 2 2e2GMv

R

6. Two identical uniform discs roll without slipping on two different surfaces AB and

CD (see figure) starting at A and C with linear speed 1v and 2v , respectively, and

always remain in contact with the surfaces, If they reach B and D with the same linear

speed and 1 3v m/s, 2v in m/s is 2( 10 / )g m s

1 3 /v m s

30 mAB

CD

27m2v

Key: 7




Sol: By law of conservation of energy,

2

2 21 AB 2 CD2

1 k 1m 1 mgh m mgh

2 R 2

2

2 21 2 CD AB2

1 km 1 mg h h

2 R

221 m 3 / 2 9 mg 32

229 4g

22 249 7m / s

( for disc 2

2

k 1R 2

)

7. 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 AB

of their

wavelength A and B at which the peaks occur in their respective radiation curves is

Key: 2

Sol: 4 2A B A BP 10 P ;R 4 10 R

2 4 4 2 4A B BR T 10 R T

2 4

4A A

B B

R T10 ;

R T

4

22 4A A

B B

T T 14 10 10

T T 2

By Wains displacement law, A BB A

T 2T




8. 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 the 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

Key: 3

Sol: final initial12.5P P100

1

final1 n

initial 0

N EP 1 1P 8 N E 2

1E Energy emitted per sec

n 3, no of half lives

SECTION-2 (ONE OR MORE THAN ONE CORRECT OPTION QUESTIONS)

This section contains 10 multiple choice questions. Each question contains four options A, B, C and D (Each question correct marking (+4), Negative mark(-2))

9. A ring of mass M and radius R is rotating with angular speed about a fixed

vertical axis passing through its centre O with two point masses each of mass 8M at

rest at O . These masses can move radially outwards along two masses rods fixed on

the ring as shown in the figure. At some instant the angular speed of the system is 89

and one of the masses is at a distance of 35

R from O . At This instant the distance of

the other mass from O is

O

A) 2

3R B) 1

3R C) 3

5R D) 4

5R




Key: D Sol: initial finalL L

2 2 2 2M 9 M 8MR MR R x8 25 8 9

2

2 2 8 1 xR R9 25 9

2

2 200 9 xR9 25 9

2

2 209 x 4RR 1 x225 9 5

10. The figures below depict two situations in which two infinitely long static line charges

of constant positive line charge density are kept parallel to each other. In there

resulting electric field, point charges q and q are kept in equilibrium between them.

The point charges are confined to move in the x direction only. If they are given a

small displacement about their equilibrium positions, then the correct statement(s) is

(are)

x

q qx

A) Both Charges execute simple harmonic motion.

B) Both charges will continue moving in the direction of their displacement.

C) Charge q executes simple harmonic motion while charge q continues moving

in the direction of its displacement.

D) Charge q executes simple harmonic motion while charges q continues moving

in the direction of its displacement.

Key: C




r O r

(1) (2)

x

E

Sol: If +q is moved rightwards, if experiences left ward force i.e. it returns to O.

Hence the equilibrium is stable.

If q is moved rightwards, it experiences a force rightwards i.e., it does not return to O

and continuous to move rightwards.

At a small displacement x from net0

1 1'O ', E2 r x r x

net0

q 1 1F E qx x2 r 1 r 1r r

0

q x x1 12 r r r

20 0

q 2x qF x2 r r r

F is leftwards, x is rightwards F x .

The particle executes SHM.

11. Two identical glass rods 1S and 2S (refractive index =1.5) have one convex end of

radius of curvature 10cm. They are placed with the curved surfaces at a distance d as

shown in the figure, with their axis (shown by the dashed line) aligned. When a point

source of light P is placed inside rod 1S on its axis at a distance of 50cm from the

curved face, the light rays emanating from it are found to be parallel to the axis inside

2S . The distance d is

SP50 cm

S P50 cm

S Pd50 cm

A) 60cm B) 70cm C) 80cm D) 90cm




Key: B Sol: R 10cm

1.5

1 1.5 1 1.5V 50 10

1 1.5 0.5V 50 10

1 0.5 1.5 2.5 1.5 V 50V 10 50 50

1.5 1 1.5 1d 50 10

1 1d 50 20

d 70

12. 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)

L LRRx

y

I / 4/ 6

A) If B

is along , ( )F L Rz B) If B

is along , 0Fx

C) If B

is along , ( )F L Ry D) If B

is along , 0Fz Key: A,B,C




Sol: If B is along x-axis eff / /B 0 , F=0 Option B is correct If B is along y-axis F BI 2 L R Option C is correct If B is along z-axis F BI 2 L R Option A is correct 13. A container of fixed volume has mixture of one mole of hydrogen and one mole of

helium in equilibrium at temperature T . Assuming the gases are ideal, the correct

statement(s) is (are)

A) The average energy per mole of the gas mixture is 2RT.

B) The ratio of speed of sound in the gas mixture to that in helium has is 6 / 5 .

C) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/2.

D) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/ 2 Key: A,B,D

Sol: (A) Total internal energy 1 2f fU nRT nRT2 2

ave per moleU 1U 5RT 3RT 2RT2n 4

(B)

1 2

1 2

1 P 2 Pmix

2 V 2 V

7R 5R1 1n C n C 32 25R 3Rn C n C 21 12 2

1 1 2 2 1 2mix1 2

n M n M M M 2 4M 3n n 2 2

Speed of sound RTV VM M

Hemix mixHe He Mix

MV 3 / 2 4 6V M 5 / 3 3 5

(D) rms rms3RT 1V VM M

,

He H

H He

V M 2 1V M 4 2




14. In an aluminum (Al) bar of square cross section, a square hole is drilled and is filled

with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are 82.7 10 m and 82.7 10 m, respectively. The electrical resistance between the

two faces P and Q of the composite bar is

Q

50 mmFeAl

7 mm

2 mm P

Q

A) 2475

64 B) 1875

64 C) 1875

49 D) 2475

132

Key: B

Sol: Al Fe

1 1 1R R R

FeAlAl Fe

AA 1

2 2 2 6

8 3

7 2 2 10 12.7 10 10 50 10

3 345 2 450 22 10 2 102.7 5 27 5

3 350 2 2562 10 2 103 5 15

315 1875R 10512 64




15. For photo-electric effect with incident photon wavelength , the stopping potential is

0V identify the correct variation(s) of 0V with and 1 / .

A)

0V

B)

0V

C) 1/

0V

D) 1/

0V

Key: A,C

Sol: 0 0hCV e w

00whC 1V

e e

01V ,

graph is in the from of y = mx c

Option C is correct.

Clearly, 0V , graph is a curve. 0V decreases with increase in . 0V becomes zero

when 0hC w

i.e. 0 (Threshold wavelength)

Option Ais correct.




16. Consider a Vernier calipers in which each 1 cm on the main scale is divided into 8

equal divisions and a screw gauge with 100 divisions on its circular scale .In the

Vernier calipers, 5 divisions of the Verinier scale coincide with 4 divisions on the

main scale and in the screw gauge, one complete rotation of the circular scale moves it

by two divisions divisions on the linear scale. Then:

A) If the pitch of the screw gauge is twice the least count of the Vernier calllipers, the

least count of the screw gauge is 0.01mm.

B) If the pitch of the screw gauge is twice the least count of the Vernier calipers , the

least count of the screw gauge is 0.005 mm.

C) If the least count of the linear scale of the screw gauge is twice the least count of

the Vernier calipers, the least count of the screw gauge is 0.01 mm.

D) If the least count of the linear scale of the screw gauge is twice the least count of

the Vernier calipers, the least count of the screw gauge id 0.005 mm.

Key: B,C

Sol: Ans: BC Vernier Callipers Screw Gauge 11MSD cm

8 Pitch of the screw 2P

1

4MSD = 5 VSD P = 2HSD 41VSD MSD

5 Least count P 1 HSD

100 50

4 1 1cm cm 1mm5 8 10

Least count = 1MSD 1VSD

1 1 1 1cm cm mm8 10 40 4

(B) 1 12HSD mm 2 1HSD mm4 4

Least count of screw Gauge 1 1 mm 0.005mm50 4

(C) L.C of linear scale of screw gauge = 1HSD

Given that 1HSD = 1 12 mm mm4 2

L.C of screw gauge 1 1 1HSD mm 0.01mm50 50 2




17. Plancks 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 is (are)

A) M c B) M G C) L h D) L G

Key: A,C,D

Sol: a b cM h c G

a b C1 2 1 1 1 3 2M ML T LT M L T

a c 2a b 3c a b 2cM L T a c 1 1

2a b 3c 0 2

a b 2c 0 3

On solving (1), (2), (3), 1 1 1a ,b ,C2 2 2

M c only A is correct

In the same way 1 3 12 2 2L h c G

L h, L G C , D are correct

18. Two independent harmonic oscillators of equal mass are oscillating about the origin

with angular frequencies 1 and 2 and have total energies 1E and 2E ,respectively.

The variations of their momenta p with positions x are shown in the figures. If

2a nb and a n

R , then the correct equation(s) is (are)

P

Energy - 1E

b

ax

P

xR

Energy - 2E

A) 1 1 2 2E E B) 22

1

n

C) 21 2 n D) 1 21 2

E E

Key: B,D




Sol: 1st Particle:

P = 0 at c = a a is the amplitude of

Oscillation 1'A ' .

At x = 0 , P = b (at mean position)

maxmv b .

maxbvm

2 2

21 max

1 m b bE mv

2 2 m 2m

1 1 1 2b b 1Am ma mn

2nd Particle:

P = 0 at x = R 2A R

At x = 0, maxRP R vm

2 2

22 max

1 m R RE mv

2 2 m 2m

2 2 2R R 1Am mR m

(B) 22 21

1/ m n1/ mn

(C) 1 2 2 2 21 1 1

mn m m n

C is not correct

(D) 2 2 2 2 2

12 2

1

E b / 2m b n a R1/ mn 2 2n 2

2 2

2

2

E R / 2m R1/ m 2

D is correct




SECTION-3

(MATRIX MATCHING TYPE QUESTIONS) This section contains 2 Matrix matching questions. Each question contains column 1 has four entries A, B, C and D and column

2 has 5 entries P,Q, R,S and T (+2 each correct match and Negative mark -1)

19. Match the nuclear process given in column I with the appropriate option(s) in column II

Column-I Column-II

A) Nuclear fusion P) Absorption of thermal neutrons by 23599 U

B) Fission in a nuclear reactor Q) 6027 Co nucleus

C) -decay R) Energy production in stars via hydrogen conversion to helium D) -ray emission S) Heavy water T) Neutrino emission Key: A RT ; B PS ; C QT ; D PQRT Sol: Conceptual 20. A particle of unit mass is moving along the x-axis under the influence of a force and its total

energy is conserved. Four possible forms of the potential energy of the particle are given in

column I ( and 0U are constants). Match the potential energies in column I to the

corresponding statement(s) in column II.

Column-I Column-II

A)

220

1 12U xU x

a

P) The force acting on the particle is zero at .x a

B)

20

2 2U xU x

a

Q) The force acting on the particle is zero

at 0.x

C)

2 20

3 exp2U x xU x

a a

R) The force acting on the particle is zero at x

D)

30

41

2 3U x xU x

a a

S) The particle experiences an attractive force towards 0x in the region x a .

T) The particle with total energy 0

4U can

oscillate about the point .x Key: A PQRT ; B QS ; C PQRS ; D PRT




Sol: A) 4 2

04 2

U x 2xU 1

2 a a

3 3

0 04 2 2 2

U 2UdU 4x 4x xF 0 x

dx 2 a a a a

F=O at both x=a,0; a P,Q,R are suitable.

at 2

0 02 2 2

2U 2Ua x a 1x ,F x 1 1

2 a a a 2 4

03UF F4a

is +ve i.e., F is not directed towards O

S is not suitable.

At x a , particle is in stable equilibrium and U=0

0 0U UTE KE4 4

At x=0, 0UU2

particle cannot reach O.

Particle executes oscillatory motion about x=-a.

B) 2

0U xU2 a

02

U xduFdx a

At x=a, 0UF 'P 'a

does not suit

At x=0, F=0 'Q ' is suitable

At x=-a, 0UF 'R 'A

is not suitable.

At x a, 'F ' is +ve when x is ve and F is ve when x is +ve the particle always

experiences force towards O S is suitable.

As F 0 at x=-a, it does not oscillate about x= -a 'T ' is not suitable.

C) 22 x

a0U xU e2 a

2 2

2 2x x2

0 a a2 2 2

UdU 2x x 2xF e e

dx 2 a a a




2

2x 2

0 a2 2

U x 2xe 22 a a

2

2x 2

0 a2 2

U x xF e 1a a

At x=a, F=0 P is suitable

At x=0, F=0 Q is suitable

At x=-a, F=0 R is suitable

F is +ve when x is ve x a and F is ve when x is +ve x a 'F ' is opposite

to x 'F' is always towards O.

'S' suitable

Between x=-a and x=0, F will be +ve. If the particle is displaced right wards of the

point x=-a i.e., towards O, it will accelerate towards O and does not return. Hence

it does not oscillate.

D) 3

03

U x 1 xU

2 a 3 a

2

03

UdU 1 xF

dx 2 a a

At x a,F 0 P and R are true

At x=0, F 0 'Q' is not suitable.

For ax ,F2

will be ve particle is not attracted towards O S is not suitable.

At x=-a, particle will be in stable equilibrium and hence it executes oscillation.




PART II : CHEMISTRY (SINGLE DIGIT INTEGER TYPE QUESTIONS)

This section contains 8 Single digit integer type questions ranging from 0 to 9, both inclusive.(Each question correct marking (+4), No Negative (0))

21. All the energy released from the reaction 0 -1, 193kJ molrX Y G is used for

oxidizing M as 3 02 , 0.25M M e E V

Under standard conditions, the number of moles of M oxidized when one mole of

X is converted to Y is 196500F C mol

Key: 4

Sol: 0 0G nFE cell

0 2 96,500 0.25G 0.5 96,500 48,250

48,250kJ

For conversion of mole of 3M M

The requirement is 48,250kJ

with 193 kJ; no. of moles of M oxidized is equal to 193/48.25 = 4

22. 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 1of water = 1.86fK K kg mol

Key: 1

Sol: f fT K m i

0.0558 1.86 0.01 i 3i one mole of complex gives 3 moles of ions complex is 3 25CO NH Cl Cl No. of Cl ions inside the coordination sphere is 1




23. The total number of stereoisomers that can exist for M is

OM

3H C

3H C3CH

Key: 2

Sol: O

No. of stereoisomers = 22 2 4n R

R

In bridge/bicyclic compounds we get less than 2n isomers

24. The number of resonance structures for N is

OH

NaOH N

Key: 9

Sol:

O - O-

-O

O

O

-

O

-

- OO

O




25. The total number of lone pairs of electrons in 2 3N O is

Key: 8

Sol: 2 3N O structure is

N N

O

..

....:

O....

O....

No. of lone pairs =8

26. 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 26Fe ]

Key: 4

Sol: 3Fe with weak field ligand SCN contains 5 unpaired e

5.9 . .B M

With strong field ligand 3Fe complex contains 1 unpaired e

1.73M

Difference is 4

27. Among the triatomic molecules/ions 2 3 2 2 3 2 2 3, , , , , , ,BeCl N N O NO O SCl ICl I and

2XeF , the total number of linear molecule(s)/ ions(s) where the hybridization of the

central atom does not have contribution from the d orbital(s) is

[Atomic number : 16, 17, 53S Cl I and 54Xe ]

Key: 4

Sol: 2 3 2 2, , ,BeCl N N O NO are linear molecules with hybridization of central atom without

d-orbitals




28. Not considering the electronic spin, the degeneracy of the second excited state 3n

of H atom is 9, while the degeneracy of the second excited state of H is

Key: 3

Sol: 2 ,2s p differ in energy is multi electron system; 2 2s p

1 to2s s first excited state

1 to2s p 2nd excited state

degeneracy of 2 3p

SECTION-2 (ONE OR MORE THAN ONE CORRECT OPTION QUESTIONS)

This section contains 10 multiple choice questions. Each question contains four options A, B, C and D (Each question correct marking (+4), Negative mark(-2))

29. The structure of D - - glucose is CHO

H OH

HO H

H OH

H OH

CH2OH The structure of L - - glucose is

A)

CHO

HO H

H OH

HO H

HO H

CH2OH

B)

CHO

H OH

HO H

H OH

HO H

CH2OH

C)

CHO

HO H

HO H

H OH

HO H

CH2OH

D)

CHO

HO H

HO H

HO H

H OH

CH2OH




Key: A Sol:

CHO

H OH

HO H

H OH

H OH

CH2OH

D-(+)-Glucose

CHO

HO H

H OH

HO H

HO H

CH2OH

L-(-)-glucose is

30. The major product of the reaction is

H3C

CH3 NH2

CO2H

2 ,0

NaNO aqueous HClC

Key: C

Sol: Use NGP of C OO

31. The correct statement(s) about 2Cr and 3Mn is (are) [Atomic numbers of 24Cr and 25Mn ] A) 2Cr is a reducing agent B) 3Mn is an oxidizing agent C) Both 2Cr and 3Mn exhibit 4d electronic configuration D) When 2Cr is used as a reducing agent, the chromium ion attains 5d electronic

configuration Key: A,B,C

A)

H3C

CH3 OH

NH2

B)

H3C

CH3 OH

CO2H

C)

H3C

CH3 OH

CO2H

D)

H3C

CH3 OH

NH2




Sol: 2Cr undergoes oxidation

2Cr is a reducing agent

3Mn undergoes reduction, acts as oxidizing agent

32. Copper is purified by electrolytic refining of blister copper. The correct statement(s)

about this process is (are)

A) Impure Cu strip is used as cathode

B) Acidified aqueous 4CuSO is used as electrolyte

C) Pure Cu deposits at cathode

D) Impurities settle as anode-mud

Key: B,C,D

Sol: In Electrolytic refining of blister copper;

1) Acidified aq. 4CuSO is an electrolyte

2) Pure copper is deposited at cathode

3) Impurities are settled as mud

33. 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

Key: A,B

Sol: 3Fe is reduced to 3Fe by

1) 2 2 /H O NaOH 2) 2 2 2/Na O H O




34. The % yield of ammonia as a function of time in the reaction

2 2 33 2 , 0N g H g NH g H at 1,P T is given below,

%yield

time

1T

If this reaction is conducted at 2, ,P T with 2 1,T T the % yield of ammonia as a

function of time is represented by

A) 1T

2T

%yi

eld

time

B)

1T

2T

%yi

eld

time

C)

1T

2T

%yi

eld

time

D) 1T

2T

%yi

eld

time Key: B Sol: Initial formation of 3NH is favoured by high temp. ( as 2N is less reactive).

Towards equilibrium (on temperature favours formation of ammonia.




35. 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) 11,4

C) 1 1,2 2

D) 1 1,4 8

Key: A

Sol: For ccp lattice; No. of atoms 4z

No. of octahedral holes 4,

No. of tetrahedral holes 8

3Al is present in 12

of the octahedral holes

2Mg are present in 18

of the tetrahedral holes




36. Compound(s) that on hydrogenation produce(s) optically inactive compound(s) is

(are)

Key: BD

Sol:

2H C

H Br

3CH2 /H Ni *

H

optically active

Br

2H C

H Br

3CH2 /H Ni

*

H

optically inactive

B)

Br

C)

3CH

BrH

2H C 2 /H Ni

3CHH Br

2CH

3CH

*

optically active

D)2CH

Br

3CH 2 /H Ni

H Br

optically inactive

A)

3H C

H Br

3CH

B)

H Br

3CH2H C

C)

H Br

3CH

2H C

3CH

D) HBr

2H C3CH




37. The major product of the following reaction is

3CH

O

O

2,. ,

LKOH H Oii H heat

A) O3CH

B) O

3CH

C) 3CH

O

D)

O3CH

Key: A

Sol: Intramolecular aldol condensation

O

O

3CH

21

( ) ,,

a KOH H Ob H Heat

O

3CH




38. In the following reaction, the major product is

3CH

2CH2H C

1 equivalent HBr

A) 2H C

Br

3CH

3CH

B)

3CH

3H C

Br

C)

3CH

2H C Br

D)

3H C

3CH

Br

Key: D

Sol: 1, 4 addition product is major

2H C

3CH

2CH 1equivalentH Br

3H C

3CH

Br




SECTION-3



39. Match the anionic species given in Column I that are present in the ore(s) given in

column II

Column-I Column-II

A) Carbonate P) Siderite

B) Sulphide Q) Malachite

C) Hydroxide R) Bauxite

D) Oxide S) Calamine

T) Argentite

Key: A PQS; B T; C QR; D - R

Sol: Siderite 3FeCO

Malachite 3 2.CuCO Cu OH

Bauxite 2 3 2.2Al O H O

Or 3 2x xAlO OH

Calamine 3ZnCO

Argentite 2Ag S




40 Match the thermodynamic processes given under Column I with the expressions

given in Column II

Column-I Column-II

A) Freezing of water at 273 K and 1 atm P) 0q

B) Expansion of 1 mol of an ideal gas into a

vacuum under isolated conditions

Q) 0w

C) Mixing of equal volumes of two ideal gases at

constant temperature and pressure in an isolated

container

R) 0syss

D) Reversible heating of 2H g at 1 atm from 300

K to 600 K, followed by reversible cooling to

300 K at 1 atm

S) 0U

T) 0G

Key: A RT; B PQS; C PQS; D - PQST

Sol: A During freezing of water entropy decreases 0syss At equation conditions; 0G

B In vacuum, free expansion work done =0

For an ideal gas in isolated condition 0 , 0q U

C For an ideal gas and isolated condition, 0, 0q U

As per first law 0W

D The process is cyclic through unique reversible path

Hence 0U (cyclic)

0G

0W (positive and negative work are same and opposite)

And 0q




PART-III: MATHEMATICS (SINGLE DIGIT INTEGER TYPE QUESTIONS) This section contains 8 Single digit integer type questions ranging from 0 to 9, both inclusive. (Each question correct marking (+4), No Negative (0)) 41. A cylindrical container is to be made from certain solid material with the following

constraints: It has a fixed inner volume of 3V mm , has a 2mm thick solid wall and is

open at the top. The bottom of the container is a solid circular disc of thickness 2mm

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 10mm, then the value of 250

V

is

Key: 4

Sol:

r

V

2

2Vhr

2 22 22 2 2Vf r r r rr

221 14 2 2V rr r

1 2 31 24 4 2 0f r V rr r

32

2V r

rr

3 1000V r

1000 4250 250

V




42. Let 2

622cos

x

x

F x t dt

for all x R and 1: 0, 0,2

f be a continuous

function. For 10,2

a , if 1 2F a is the area of the region bounded by

0, 0,x y y f x and ,x a then 0f is

Key: 3

Sol: 1 11 110

2 0 0a

f x dx F a f a F a f F

1 2 2 24 cos 2cos6

F x x x x

11 2 2 24cos 4 sin 2 .2 2sin 26 3

F x x x x x x

11 30 4. 34

F

43. The number of distinct solutions of the equation

4 4 6 65 cos 2 cos sin cos sin 24

x x x x x

in the interval 0,2 is

Key: 8

Sol: Sol: 2 2 2 2 25 cos 2 1 2cos sin 1 3sin cos 24

x x x x x

2 25 5cos 2 sin 2 04 4

x x

2 2sin 2 cos 2x x

cos4 0x

3 5 7 9 11 13 15 174 , , , , , , ,2 2 2 2 2 2 2 2 2

x

3 5 7 9 11 13 15, , , , , , ,8 8 8 8 8 8 8 8

x

8 solutions in 4 2x




44. Let the curve C be the mirror image of the parabola 2 4y x with respect to the line

4 0x y . If A and B are the points of intersection of C with the line 5y , then

the distance between A and B is

Key: 4

Sol: Curve c is 2 4 4x a y

24 4 4 ... 1x y

5..... 2y 24 4x

4 2x

P.I of 1 & 2

2, 5 6, 5A B

AB=4

45. The minimum number of times a fair coin needs to be tossed, so that probability of

getting at least two heads is at least 0.96, is

Key: 8

Sol: 11 1 0.962n

n

10.042n

n

2 25 1n n

8n




46. 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 mn

is

Key: 5

Sol: 6!5!n

6 25! 5!m P

5!5!30 55!6!

mn

47. If the normals of the parabola 2 4y x drawn at the end points of its latus rectum are

tangents to the circle 2 2 23 2x y r . Then the value of 2r is Key: 2 Sol:

(1,2)

(1,-2)

Equation of the normal at the end points of latus rectum are 2 1 3y x x 2 1 3y x x 3 0;x y 3 0x y Centre of the circle = 3, 2

3 2 3

22

r

3 2 3 1

2 2r

2 2r




48. Let :f R R be a function defined by , 2

0, 2x x

f xx

where x is the greatest

integer less than or equal to x . If

22

1 2 1xf x

I dxf x

, then the value of 4 1I is

Key: 0

Sol: 2 2

2

2

2

0 2

x xf x

x

1 1 2, 1

10 1 2, 1

x x xf x

x x

2 1x 2 2x 1 0x 2 1x

0 1x 2 0x 1 2x 2 1x

1 0x 1 0f x 0 1x 1f x

1 2x 1 0f x

22

1 2 1xf x

I dxf x

2 2 2 20 1 2 2

1 0 1 22 1 2 1 2 1 2 1xf x xf x xf x xf x

dx dx dx dxf x f x f x f x

2

1

0 0 02xdx

22

1

1 1 1. 2 12 2 4 4

x

4 1 0I




SECTION-2

(ONE OR MORE THAN ONE CORRECT OPTION QUESTIONS) This section contains 10 multiple choice questions. Each question contains four options A, B, C and D (Each question correct

marking (+4), Negative mark(-2))

49. 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 0P x y z and

2 : 2 1 0P x y z . Let M be the locus of the feet of the perpendiculars drawn from

the points on L to the plane 1P . Which of the following points lie(s) on M?

A) 5 20, ,6 3

B) 1 1 1, ,6 3 6

C) 5 1,0,6 6

D) 1 2,0,3 3

Key: A,B,C,D

Sol: L is any line passing through origin and in the plane 3 0x y L has infinite

directions are consists of all the points in the plane 3 0x y

The locus of feet of perpendicular from the pints on L to the plane 1P is the plane 1P

it self

All A, B, C, D satisfy 1P

50. Let P and Q be distinct points on the parabola 2 2y 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

Key: A,D




Sol: 21 1,2P at at 22 2,2Q at at

4 2a 12

a

2 21 2 1 22 2 0x at x at y at y at Passing through (0,0) 2 2 2 21 2 1 24 0a t t a t t 2 1 2 1 2 4 0a t t t t

21

4tt

3 2OPQ

2 21 2 1 21 2 2 3 22

at at at at

2

1 21 2

2 3 22

a t t t t

1 2 3 2t t

11

4 3 2tt

11

4 3 2tt

11

4 3 2tt

21 13 2 4 0t t 2

1 13 2 4 0t t

21 1 12 2 2 4 0t t t

1 1 12 2 2 2 2 0t t t 1 2 2 2t or 21 1,2 1, 2 4,2 2at at or

51. Let y x be a solution of the differential equation 11 1x xe y ye . If 0 2y , then which of the following statements is (are) true?

A) 4 0y B) 2 0y C) y x has a critical point in the interval 1,0 D) y x has no critical point in the interval 1,0 Key: A, C




Sol: 11 1

x

x xdy e ydx e e

1. 1x

xe

xeI R e e

1 xy e x c

, 0,2x y

2 2 0 c

4c

1 4xy e x

41 xxy

e

4 0y

21 4

1

x x

x

e x edydx e

2

1 4

1

x x x

x

e xe ee

1 3 0x xxe e

1 3 0xx e

13

xex

has a solution in 1,0

11,e

0,1

10,3




52. Consider the family of all circles whose centres lie on the straight line y x if this

family of circles is represented by the differential equation 11 1 1 0Py Qy where

,P Q are functions of ,x y and 1y (here 2

1 112,

dy d yy ydx dx

), then which of the

following statements is (are) true?

A) P y x B) P y x

C) 211P Q x y y D) 21 1P Q x y y y Key: B, C

Sol: 2 2 2 2 0x y ax ay c

1 12 2 2 2 0x yy a ay

1

11x yya

y

Diff. w.r.t x

21 11 1 1 111 1 0y yy y x yy y

211 1 1 11 1 0y x y y y y Comparing with G. E.

;P y x 21 11Q y y

21 11P Q x y y y




53. Let :g R R be a differentiable function with 10 0, 0 0g g and 1 1 0g . Let

, 0

0, 0

x g x xxf x

x

And xh x e for all x R . Let foh x denote f h x and hof x 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 h is differentiable at 0x D) h f is differentiable at 0x

Key: A,D

Sol: A) 10

limx

f x f xf x

x

0

0limx

x g xx

x

0

limx

g xx

1

1

0

00 lim 0

1xg x g

fx

1

1

0

00 lim 0

1xg x g

fx

f is differentiable at 0x

B) , 0xh x e x

, 0xe x




1 , 0xh x e x

, 0xe x

1 10 1, 0 1h h

f is not differentiable at 0x

C) ,xfoh x f e not defined at x=0

not differentiable at x=0

D) , 0hof x h g x x

, 0h g x x

1, 0x

, 0g xhof x e x

, 0g xe x

1, 0x

1 1g xhof x e g x

10, 0 0g

differentiable at 0x




54. Let sin sin sin6 2

f x x

for all x R and sin2

g x x for all x R . Let

f g x denote f g x and g 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 g is 1 1,2 2

C) 0

lim6x

f xg x

D) There is an x R such that 1g f x

Key: A,B,C

Sol: sin sin sin6 2

f x x

1 sin sin 12

x

sin sin6 6 2 6

x

1 1sin sin sin2 6 2 2

x

0

sin sin sin6 2lim

sin2

x

x

x

0

cos sin sin cos sin cos6 2 6 2 2lim

cos2

x

x x x

x

sin sin sin sin sin2 6 2 2

f g x f x x




55. Let PQR be a triangle. Let a QR , b RP

and c PQ

. If 12, 4 3a b and

. 24b c , then which of the following is(are) true?

A) 2

122c

a B)

2

302c

a

C) 48 3a b c a D) . 72a b

Key: A,C,D

Sol: 0a b c b c a

2 2b c a 2 2 22 .b c b c a

248 48 144c

2 48c

A)

248 12 12

2 2c

a

B) 2

48 12 362 2c

a

C) b c a

0a b a c

a b c a

2 sina b g a b

12 12 4 32

48 3 D) a b c

22 22 .a b a b c

. 72a b

cos 3 / 2




56. 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 3Y Z Z Y B) 44 44X Y

C) 4 3 3 4X Z Z X D) 23 23X Y

Key: C, D

Sol: Given ,Tx x ,Ty y Tz z

3 4 4 3 3 4 34T T Ty z z y y z y

4 3 3 44 4T T T Tz z

3 3 44 4 4z z

4 3 3 4 4 3 3 4z y y z z y y z Not skew symmetric

44 44 44 4444 44 44 444T T Tx y x y x x y Symmetric

3 4 4 34 3 3 4 T T T T Tx z z x z x x z

4 4 33z x x z

3 4 4 3z x x z

4 3 3 4x z z x

Skew symmetric

23 23 23 2323 23 T T Tx y x y x y

23 23 23 23x y x y

Skew - symmetric




57. Which of the following values of satisfy the equation

2 2 2

2 2 2

2 2

1 1 2 1 3

2 2 2 2 3 648

3 3 2 3 3

?

A) 4 B) 9 C) 9 D) 4 Key: B,C

Sol:

2 2 2

2 2 2

2 2 2

1 1 2 1 3

2 2 2 2 3 648

3 3 2 3 3

2 2 1 3 3 2,.R R R R R R

2 2 21 1 2 1 33 2 3 4 3 6 6485 2 5 4 5 6

3 3 2R R R

2 2 21 1 2 1 33 2 3 4 3 6 648

2 2 2

2 2 21 1 2 1 33 2 3 4 3 6 324

1 1 1

2 1 3 3 2;C C C C C

2 2 21 2 3 2 53 2 2 2 324

1 0 0

2 21 2 3 23 2 2 0 324

1 0 0

22 2 324

2324 81 9

4




58. In 3,R consider the planes 1 : 0P y and 2 : 1P x z . Let 3P be a plane, different

from 1P and 2P , which passes through the intersection of 1P and 2P . If the distance of

the point 0,1,0 from 3P is 1 and the distance of a point , , from 3P is 2, then

which of the following relations is(are) true?

A) 2 2 2 0 B) 2 2 4 0

C) 2 2 10 0 D) 2 2 8 0

Key: B,D

Sol: Eg of the plane is

1 0x z xy

Distance from 0,1,0 is 1

2

0 0 1 11

1 1

2 21 2

2 21 2 2

12 12

Eg of the plane is 1 02yx z

2 27 2 0x y

Distance from , , is 2

2 2 2

21 4 4

2 2 2 6 8 4or




SECTION-3



59.

Column-I Column-II

(A) In 2R , if the magnitude of the projection vector

Of the vector i j on 3i j is 3 and if

2 3 ,

then possible value(s) of is (are)

(P) 1

(B) Let a and b be real numbers such that the

function 2

2

3 2, 1, 1

ax xf x

bx a x

is

differentiable for all x R . Then possible

value(s) of a is(are)

(Q) 2

(C) Let 1 be a complex cube root of unity. If

4 3 4 32 23 3 2 2 3 3n n

4 323 2 3 0n , then possible value(s) of n is(are)

(R) 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 ,5, ,a q b is an arithmetic

progression, then the value(s) of q a is(are)

(S) 4

(T) 5

Key: A P,Q; B P, Q; C P, Q, S, T; D Q, T




Sol: A) 3 32

3 2 3 3 2

3 3 2 3

4 0

0

2

3 2 3

3 3 2 3

4 4 3

3

3 3 3 2 3

3 3

1

2 1or

A P,Q

B) 2

2

3 2 11

ax xf xbx a x

1 1x x

Lt f x Lt f x

( f is diff. x R )

23 2a b a

2 3 2a a b

16 1

1ax x

f xb x

1 11 1f f 6a b

2 3 2 6a a a




2 3 2 0a a 1( )2a or

B PQ

C) 4 3 4 3 4 32 2 23 3 2 2 3 3 3 2 3 0n n n

2 23 3 2 2 1 1 3 5 1

2 2 22 3 3 2 1 3 5 5 1

2 2 2 2 23 2 3 3 2 5 5 1

4 34 3 4 3 25 1 1 0nn n 1, 2, 4, 5n n n n

C PQST

D) 2 4 2ab ab a ba b

22

aba

,5, ,a q b -A.P

5 5a d d a

232

ab a da

23 52

aa aa

22 17 30 0a a 6a or 5 / 2 2 2 5q a d a a 10 a 6, 4a q 5 / 2a , 15 / 2q 2,5q a

,D Q T




60.

Column-I Column-II

(A) In a triangle XYZ ,let a,b,and c be the lengths of the sides opposite to the angles X,Y and Z respectively. If

2 2 2 sin2sin

X Ya b c and

Z

,then

possible value(s) of n for which cos , 0n is (are)

(P) 1

(B) In a triangle XYZ let a,b,and c be the lengths of the sides opposite to the angles X,Y and Z respectively. If 1 cos2 2cos2 2sin sin ,X Y X Y then

possible value(s) of ab

is (are)

(Q) 2

(C) In 2R let 3 , 3i j i j and 1i j be

the position vectors of X,Yand Z with respect

to the origin O, respectively.If the distance of

Z from the bisector of acute angle of OX

with

OY

is 32

then possible value(s) of is (are)

(R) 3

(D) Suppose that F denotes the area of the region bounded by

20, 2, 4 1 2x x y x and y x x x ,where 0,1 ,Then the value(s) of

8 23

F ,when 0 and 1 is (are)

(S) 5

(T) 6

Key: A PRS; B P; C PQ; D ST




Sol: (A) 2 2 22 a b c 2 2 22 sin sin sinx y z 22sin sin sinx y x y z

sin 1sin 2

x yz

12

cos 0 cos 02

nn

Is odd A matches to P,R,S (B) 1 cos2 2cos2 2sin sinx y x y 2 21 1 2sin 2 1 2sin 2sin sinx y x y 2 2sin sin sin 2sin 0x x y y

sin 2 1sin

x ory

But sin 1sin

xy only possibility

B matches to P (C) Bisectors of &OX OY

one x y or 0x y

Distance from ,1z to 0x y

Is 2 1 3 1 22 2

or

1 2or C matches to P,Q Other equation 0x y dose not give solution for B P.T.O (D)

x=2x=0

y=3




when 0 3a y

We have 0

82 23

i xdx

2080 6 2 6 23

F xdx

80 2 63

F

When 1 1 2a y x x x

3 0,11 1,2x if x

x if x

y=3-x y=1+

x

1 2 10 1 01 3 1 2F x dx x dx xdx

81 2 53

F

,Q S T

Jee Adv 2015 Paper 1

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Transcript of Jee Adv 2015 Paper 1