Test RIDER Advanced - m.media-amazon.com · A uniform ring of mass m is lying at a distance 3 a...

PRACTICE SET 5 273

A to the Various IITs...S ingle Door Ent r y

Advanced

Paper 1

Duration : 3 Hours Max. Marks : 180

Test RIDER

Pra

cti

ce

Se

t5

Question Booklet Code 4

Please read the instructions carefully. You are allotted 5 minutes speciallyfor this purpose.

Question Paper Format

Marking Scheme

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This booklet is your question paper. Attempt all the questions.

Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and

electronic gadgets are not allowed.

Write your name and roll number in the space provided on the bottom of this page.

The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists

of three sections.

contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d)

out of which only one is correct.

contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d)

out of which one or more than one is/are correct.

contains 5 questions. The answer to each question is a single–digit integer, ranging

from 0 to 9 (both inclusive)

For each question in Section 1, you will be awarded for correct answer and zero mark for

unattempted. No negative marks will be awarded for incorrect answers in this section.

For each question in Section 2, you will be awarded for correct answer(s) and zero mark for

unattempted questions. In all other cases, minus one (–1) mark will be awarded.

For each question in Section 3, you will be awarded for the correct answer and zero mark for

unattempted questions. In all other cases, minus one (–1) mark will be awarded.

Section 1

Section 2

Section 3

2 marks

4 marks

4 marks

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

16pb.

IBL

pc.

IBL

2pd.

IBL

4p

2. A glass prism of refractive index 1.5 is immersed in water (refractive index is 4/3). A light beam

incident normally on the face AB is totally reflected to reach the face BC if

a. sin /q > 8 9 b. 2 3 8 9/ sin /< <qc. sin /q < 2 8 d. None of these

3. The variation of induced emf (e) with time (t) in a coil if a short bar magnet is moved along its

axis with a constant velocity is best represented as

AdvancedTest RIDER

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

B A

θ

C

Part ISection 1 Single Correct Option Type

10

a.

e ee e

b. c. d.

4. A uniform ring of massm is lying at a distance 3 a from centre of a sphere of

mass M just over the sphere where a small radius of ring as well as that of

sphere. Then, gravitational force exerted is

a.GMm

a8 2b.

GMm

a3 2

c.3

2

GMm

ad.

3

8 2

GMm

a

5. A small block is shot into each of the four tracks as shown in the figure 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

6. The spring mass system inside frictionless tube rotating with constant w as

given. Find angular frequency wn of mass for SHM about its mean position.

a.k

mb. w c.

k

m- w2

d.w

2k

7. An optical system consists of a diverging and converging lens. Focal length of lens is 40 cm.

Object is at a distance of 80 cm in front of converging lens. Distance of image from converging

lens is

a. 40 cm b. 100 cm

c. 120 cm d. 80 cm

8. The intensity of X-rays from a coolidge tube is plotted against

wavelength l as shown in the figure. The minimum wavelength

found is lc and the wavelength of the K a line is lk . As the

accelerating voltage is increased.

a. l lk c- increases b. lk decreases

c. l lk c- decreases d. lk increases

9. Consider the given system of a ball of mass m, massless rod and

two springs. Find ratio of L a/ 2 for which system will become

unstable.

a.k

gb.

k

mg

c.2k

mgd.

k

mg2

PRACTICE SET 5 275

Very small80 cm

O

m

ω

k

v v

a. b.

v v

c. d.

λc λk

λ

I

m

k kL

a

a

a

√3a

10. Two equal point charges are fixed at x a= - and x a= + on the x-axis. Another point charge Q is

placed at origin. The change in electrostatic potential energy of the system when Q is displaced

by a small distance x along the x-axis is approximately proportional to

a. x b. x2c. x3

d.1

x

11. If the tension in a stretched string fixed at both ends is changed by 20%, the fundamental

frequency is found to increase by 15 Hz. Then, the

a. original frequency is 157 Hz

b. velocity of propagation of transverse wave along the string changes by 5%

c. velocity of propagation of transverse wave along string changes by 10%

d. fundamental wavelength of string does not change

12. A point isotropic source of sound power 1 mW emits sound of frequency 170 Hz in all directions.

Velocity of sound is 340 m/s, then

a. intensity at any point at distance r depends on 1 2/ .r

b. at r = 250

pm, the loudness of source is 60 dB

c. amplitude of sound wave depends on 1/ r

d. amplitude of oscillation of a point at 4 m is 2A and at 55 m it is - 8

55

Aat t = 0

13. A spherical hole is made in a solid sphere of radius. The mass of the sphere before

hollowing was M 0 . The gravitational field at the centre of the hole due to the

remaining mass is

a.GM

a

0210

b. zero c.GM

a

02

d.GM

a

022

14. A golf club hit a golf ball and that collision last for only Dt. Mass of golf ball is 45 g and maximum

range corresponding to given shot is 160 m. [Take g = 10 m/s2] [Distance covered by ball during

collision is 2 cm].

a. impulse imparted by golf club is 1.8 N-s

b. average force applied is 1800 N

c. estimated time of collision is 0.001 s

d. average acceleration of ball during collision is 40000 m/s 2

15. A long straight wire along the z-axis carries a current i in the negative z-direction. The magnetic

field B at a point having coordinates ( , )x y on the z = 0 plane is

a.m

p0

2 22

I x y

x y

( $ $ )

( )

i j-+

b.mp0

2 22

I x y

x y

( $ $)

( )

j i-+

c.m

p0

2 22

I x y

x y

( $ $)

( )

i + j

+d.

mp0

2 22

I y x

x y

( $ $)

( )

i j-+

276 AdvancedTest RIDER

Section 2 More Than One Correct Option

This section contains . Each question has four choices, (a), (b), (c)

and (d) out of which or is/are correct.

5 multiple choice questions

one more than one

a

16. A spring of force constant k = 300 N/m connects two blocks having masses 2 kg and 3 kg, lying

on a smooth horizontal plane. If the spring block system is released from a stretched position, the

number of complete oscillations in 1 min is 6n. Find the value of n. Take p = 10.

17. A steel bar of length L is held between rigid supports and heated non-uniformly in such a

manner that the temperature increases DT at distance x from one end is given by DT T x L= 02 2/

as shown in the figure.

The stress in the bar, (assume that modulus of elasticity for steel is E and thermal expansion

coefficient is a) is given as s a= E T

x

0 . Find the value of x .

18. When the voltage applied to an X-ray tube is increased from 10 kV to 20 kV, the wavelength

interval between the K a line and the short wave cut off of the continuous X-ray spectrum

increases by a factor 3. Find the atomic number of element of which the tube anode is made. Fill

second digit (tenth) of your answer.

19. Light incidence at an angle q w.r.t. the axis on one plane end, of a transparent cylindrical fiber of

refractive index n = 1.25. Determine the maximum value of q, so that the light entering the rod

does not come out of curved surface. Answer the value q/ .5

20. A parallel beam of nitrogen molecules moving with velocity v = 400 m/s impinges on a wall at an

angle q = °30 to its normal. The concentration of molecules in the beam n = ´ - -0.9 10 cm19 3 . Find

the pressure exerted by the beam on the wall assuming the molecules to scatter in accordance

with the perfectly elastic collision law.

x

L

x

∆TTemperature

θ

Section 3 Integer Answer Type

This section contains . The answer to each question is a single-digit

integer, ranging from to (both inclusive).


0 9

Part II

21. Among the hydroxides of period 3 elements, the element whose oxide is amphoteric has value of

quantum numbers are

a. n l s= = =3 1 1 2, , / b. n l s= = =3 1 3 2, , / c. n l s= = =3 1 1, , d. n l s= = =3 0 1 2, , /

22. An aromatic organic compound A having molecular mass 112.5 containing C, H and chlorine

only, have 4 degree of unsaturation what will be the product when A undergoes reaction with

sodium hydroxide?

23. x mole KIO3 is treated with excess of KI, liberated I2 which was dissolved in freshly prepared

starch solution. Which was neutralised by 60 mL 0.1 N Na S O2 2 3 until white ppt. was obtained.

What is x?

a. 10 3- mol b. 10 5- mol c. 5 mol d. 6 mol

24. Some oxides of p-block elements show colour while some do not among following, which of the

following pair of oxides are coloured?

a. NO 2 and N O2 3 b. NO 2 and N O2 5 c. N O2 4 and NO 2 d. NO and N O2

25. 1 mol of an ideal gas A ( )C RV = 3 and 2 moles of an ideal gas B C RV =æèç

öø÷

3

2taken in a container

and expanded reversibly and adiabatically from 1 L to 4 L starting from initial temperature of320 K.

DE for the process is

a. - 240R b. 240R c. 480R d. - 960R

26. CH CH CH CH CBrCl3 2 2 3

C H CO) O6 5 2 2¾ ¾ == + ¾¾¾¾®(

M

M will be

a. CH CH C H

Cl

CH CBrCl3 2 2 2¾½

¾ ¾— b. CH CH C H

CCl

CH Br3 2

3

2¾ ¾½

¾ ¾

c. CH CH C H

Br

CH CCl3 2 2 3¾ ¾½

¾ ¾ d. CH CH C H CH

CBrCl

Cl3 2 2

2

¾ ¾½

¾ ¾

27. What is the magnetic moment of coordination compound formed during brown ring test?

a. 3.87 BM b. 4.92 BM c. 5.92 BM d. 2.83 BM

AdvancedTest RIDER

Section 1 Single Correct Option Type

10

a. b. c. d.

OH Cl

OH

Cl

OH

None of these

28. Which of the following will produce a stable non-planar carbocation?

a. Addition of antimonypentafluoride to cycloheptatrienyl fluoride

b. Addition of bromonium ion to propene

c. Addition to acid to propene

d. Addition of argentium ion to triphenylmethyl chloride

29. Which of the following is optically inactive?

a. Trioxalato chromate (III)

b. Trans-dichlorobis ethylenediamine platinium (II) chloride

c. Trans-diamine dichlorobis ethylenediamine chromium (I)

d. Both a and c.

30. Derived name of iso-octane is

a. dimethyl n-pentyl methane b. iso-butyl trimethyl methane

c. trimethyl iso-butyl methane d. iso-propyl trimethyl methane

31. Decomposition of 3 2 2A g B g C g( ) ( ) ( )¾® + follows first order kinetics. Initially, only A is

present in the container pressure developed after 20 min and infinite time are 3.5 atm and 4 atm

respectively. Which of the following is true?

a. t50 20% = min b. t75 40% = min c. t99 64 3% /= min d. t87.5% = 60 min

32. Choose that which of the following statement (s) is/are correct regarding given equations

written below?

A. FeO SiO Fe SiO2 2 3+ ¾®B. ZnO C Zn CO+ ¾® +C. 2Al O C 4Al 3CO2 3 2+ ¾® +3

a. A represents formation of slag and B represents reduction of zinc

b. A represents formation of flux and B represents reduction of zinc

c. A represents formation of slag and C represents Hall Heroult process

d. B represents reduction of zinc and C represents Hall Heroult process

33. Which of the following reactions complete with racemisation?

PRACTICE SET 5 279





only one more than one

a.

c.

b.

d.

Br2

UV

H , H O⊕2

CHOHCN

Basic medium

C—NH2Br + KOH2

O

34. Out of the following the correct statement(s) is/are

a. the structure of diamond and corrundum is same.

b. a mixture of 5-10%CO2 andO2 is known as carbogen which is used for artificial respiration in pneumonia

patient.

c. SnCl2 is a strong oxidising agent.

d. PbO is a yellow orange coloured powder commonly known as litharge.

35.

H

Me

H

Me

Me

H

H

Me== ¾® ==

(cis) (trans)

Correct sequence of reagents used for above conversion are

a. (i) H H O2Å, (ii) H /Å D b. (i) Ni (ii) Br H2 / (ii) alco KOH D

c. (i) Br /CCl2 4 (ii) 2NaNH2, D (iii) liq.NH , Na3 d. (i) Br /CCl2 4 (ii) NaI + Acetone, D

36. Consider the equation A g B g C g( ) ( ) ( )+ 2 s when the reaction was carried out at 120°C, the

equilibrium concentrations of A and B were 3 M and 4 M respectively. When the volume of the

vessel was doubled and system is allowed to reach equilibrium, the concentration of B was found

to be 3 M. The original concentration of C will be

37. Electrons in a sample of H-atom make transition from state n x= to some lower excited state.

The emission spectrum from the sample is found to contain only the lines belonging to a

particular series. If one of the photons had energy of 0.6375 eV. Then, find the value of x .

(Take 0.6375 eV 0.85 e V= ´3

4)

38. What will be the change in oxidation state during transformation of KMnO4 to manganese

dioxide?

39. Oxidation state of Ti in Zieglar Natta catalyst is

40. In the acid-base titration [ ]H PO (0.1M) NaOH(0.1 M)3 4 + emf of the solution is measured by

coupling this electrodes with suitable reference electrodes. When alkali is added, pH of the

solution is in accordance with equation.

E Eo ocell cell 0.059= ´ pH

For H PO3 4 K K K13

28

31310 10 10= = =- - -, ,

What is the cell emf at the 2nd end point of titration if E cell° at the this stage is 1.3805 V?




integer, ranging from to


0 9 (both inclusive)

Part III

41. P x y z1 2 2 0º - + = and P x y z2 2 4º - + = are the equations of two given planes. A plane P

passes through the point ( , , )1 2 1- and is perpendicular to the two given planes P1 and P2 . If the

distance of the point (1, 2, 2) from the plane P is equal to k 2, then the value of k is

a. 2 b. 3 c. 1 d. 4

42. Two media drawn from acute angles of a right angled triangle intersect at an angle p /6. If the

length of the hypotenuse of the triangle is 3 units, then the area of the triangle (in sq units) is

a. 3 b. 3 c. 2 d. 9

43. The equation x x a3 3 0- + =[ ] (where [.] denotes the greatest integer function) will have three

real and distinct roots if

a. a Î - ¥( , )2 b. a Î( , )0 2

c. a Î - ¥ - È ¥( , ) ( , )2 0 d. a Î -[ , )1 2

44. If f and its derivatives be differentiable everywhere and f f¢ = =( ) ( )2014 20141

2014

f x dx( ) ,=ò 10

2014then value of x f x2

0

2014¢¢ò ( )dx will be

a. 2014 b. 1007 c. 2008 d. 2016

45. Let x x x1 2 32 2 2 2 2 2= = + = + +, , ……, x n = + + +2 2 2 ……n (times). Then,

2 11

2

1

4

1

8

1

245

2005sin + + + +¼+æ

èçöø÷

° must be

a. x2007 b. x2006 c. x2005 d. None of these

46. If a a a a0 1 2 1, , ¼ -n are n th roots of unity where a p pk

k

n

i k

nk n= + £ £ -cos sin

,2 2

0 1, then the

value of 22 11n

n n

n

n

- æèç

öø÷

æèç

öø÷

¼ -æèç

öø÷

sin sin sin( )p p p

equals

a. n b. -1 c. n - 1 d. n/3

47. The straight lines L L L1 2 3, , are parallel and lie in the same plane. A total number of m points on

L1 , n points on L2 and k points on L3 , then the maximum number of triangles formed with

vertices at these points are

a.m n k C+ +

3 b.m n k m n kC C C C+ + - - -3 3 3 3

c.m n kC C C3 33

+ + d.m n k C+ +

3

Section 1 Single Correct Option Type

10

48. If |sin | cos sin | |2 2 2 2 226 2 25x x x x x+ - = + + - , then x lies in

a. [ , ]- 6 6 b. [ , ]-7 7 c. [ , ]-5 5 d. None of these

49. If three squares are selected at random from chessboard, then the probability that they form the

letter L is

a.49 64

3/ C b.

196 643

/ C c.36 64

3/ C d.

98 643

/ C

50. x xdy

dxx y

f xy

f xy( ) ( )

( )

( ),+ + + =

¢1 1 where x ¹ - 1, then f xy( ) is equal to

a. f xy c x y( ) ( )= - b. f xycx

y( ) = c. f xy c x( ) ( )= + 1 d. f xy c x( ) ( )= - 1

51. Let u and v be unit vectors inclined at an angle q such that for some vector w?

w w u v+ ´ =If b =[ ]u v w then,

a. v w× = cos q b. b q= 1 2 2/ sin c. u w× = cos q d. u w× = -1 b

52. A straight line through the vertex P of a triangle PQR intersect the side QR at the point S and the

circumcircle of the triangle PQR at point T . If S is not the centre of the circumcircle, then

a.1 1 2

PS ST QS SR+ <

´b.

1 1 2

PS ST QS SR+ >

´

c.1 1 4

PS ST QR+ < d.

1 1 4

PS ST QR+ >

53. Let f R R: ® be a function defined by f xf x

f xx R( )

( )

( )+ = -

-" Î1

5

3. Then, which of the following

statement (s) is/are true?

a. f f( ) ( )2008 2004= b. f f( ) ( )2006 2010= c. f f( ) ( )2006 2002= d. f f( ) ( )2006 2018=

54. Two sides of rhombus OABC (lying entirely in first quadrant or third quadrant) are yx=3

and

y x= 3 . If the area of rhombus OABC is equal to 2sq units. Then, the possible coordinates of B

is/are (where O being the origin)

a. ( , )1 3 1 3+ + b. ( , )- - - -1 3 1 3

c. ( , )3 1 3 1- - d. None of these

55. CF is the internal bisector of angle C of DABC, then CF is equal to

a.2

2ab

a bC

+cos ( / ) b.

a b

abC

+2

2cos ( / )

c.b A

B C

sin

sin ( / )+ 2d. None of these






only one more than one

56. Find the value of ( ) [ ]x d x2

0

21+ò , where [ ]x represent GIf

57. If the triangle ABC whose vertices are A B( , , ) ( , , )- -1 1 1 1 1 1 and C ( , , )1 1 1- is

projected on xy-plane. Then, the area of the projected triangle is

58. The locus of the centre of a variable circle touching two circles of radius r r1 2and externally,

which also touch each other externally is a conic. Ifr

r

1

2

7

5= and the eccentricity of the conic is k,

then k is equal to ……

59. Let a b c, , be three unit vectors such that a is perpendicular to the plane of b and c. If the angle

between b and c is p /3 then, | |a b a c´ - ´ =

60. Number of solutions of the equation |sin cos | tan cot , [ , ]x x x x x+ + + = Î2 3 0 62 2 p are



integer, ranging from to


0 9 (both inclusive)

1. (b) Idea When a current carrying wire is placed in amagnetic force I ( )L B× will act on it. Where Iis current, L is length and B is magnetic field.By symmetry, in this question the shape willbe circular.

Consider a small arc as below

Clearly, net force 2T dsin ( )θ will be by magnetic forceon the wire hence,

2 90T d F I dLBmsin( ) sinθ = = °

2T d I dLB( )θ ≈

TIdLB

dIB= =

2 θ

= = ×IBR IBL

2π( )QL R= 2π

TIBL=2π

TEST Edge Questions are framed very often in JEEAdvanced, on this concept wire could be of anyshape. The wire may be moving also. The pointsshould be kept in mind.

(i) The net force on the circular wire will be zero.(ii) To find tension we have to consider a small arc

element.

2. (b) Idea The problem is based on Snell’s law andconcept of TIR is used. As the ray incidentnormal to the face AB, hence the ray will passundeviated and will incident on side BC asshown in figure and there total internalreflection will take place.

Consider the diagram below

Clearly, the angle of incidence at the face BC will alsobe θ as shown in above diagram.

Now for total internal reflectioni c> θ

sin sinθ θ> c

sinθ µµ

> w

g( sin sin )Qµ θ µg w= °90

sin//

θ > =4 33 2

89

TEST Edge In every year, questions are asked in JEEAdvanced based on this concept. Variety ofquestions can be framed by varying the angle ofincidence of light as well as varying refractiveindices.

Note (i) For total internal reflection to take place lightray must travel from denser to rarer medium.

(ii) At just critical angle, refractive angle will be 90°(iii) At angle greater than critical angle then total

internal reflection will take place.

3. (d) Idea The problem is asked about the bar magnetis moved towards the coil then flux linkedwith the coil is changing, hence emf will beinduced according to Faraday’s law.

Ed

dtind = − φ

When the magnet is leaving the coil then fluxdecreases hence current flow tend to flow inopposite direction.

When the bar magnet is entering and leaving the coil,the polarity of emf should be opposite in two cases.

Only in option (b) polarity is different.

TEST Edge Questions are framed in JEEAdvanced by changing the magnetic flux bydifferent modes like by rotation, by translation,combination of movement of both coil and magnetor by the variation of the field. The points should bekept in mind

(i) Variation of flux (by any means) is necessary forthe emf to be induced.

(ii) For induced current circuit must not beincomplete.

4. (d) Idea To solve this problem we have to considerthe field at axial point of the ring i.e.,

EGmx

R x= −

+( ) /2 2 3 2, field will be directed

towards the centre of the ring.

We will treat sphere as a point mass placed atits centre.

For all point out the sphere we can treat it as pointmass at its centre so, effectively it will be the forcebetween ring and point mass.


dθ

dθdθ

dθ

T

Fm

A B

θ

i=θ

C

i

Gravitation field at x a= 3 on axis of ring is

EGm x

R x

Gm a

a a= −

+= −

+( )( )

[ ]/ /2 2 3 2 2 2 3 23

3

= − 38 2

Gm

a

F MEGMm

a= = − 3

8 2

TEST Edge Questions can be framed by usingdifferent configurations. In place of ring it can beother shaped objects like shell, disc etc. The pointsshould be kept in mind

(i) We should calculate the field at axial points.(ii) We should always be aware about symmetry.

5. (b) Idea To solve these type of questions, the conceptof circular motion will be used. We will drawthe FBD of the system and then we will writeequation according to net centripetal force. Atthe highest point we will have

N mgmv

R+ =

2

As the block rises to same height in all cases, hencespeed of the highest point will be same at all pointsaccording to conservation of energy.

We can write

N mgmv

R+ = 0

2

Nmv

Rmg= −0

2

Nmv

R mmgmax

min= −0

2

R is minimum for first case hence (b) is correct.

TEST Edge These types of questions are generallyasked in JEE Advanced IIJ-JEE. Questions could beframed by varying the radius of curvature of thetrack. The points should be kept in mind

(i) We should be careful in applying the directionof normal force acting on the system.

(ii) Sometimes the block will leave the track in thatcase we have to put N = 0 .

6. (b) Idea We will use the concept of centripetal forceto solve the question. The required force forthe circular motion will be provided by thespring i.e., when elongation in the spring is xthen,

kx m L= ω2

The free body diagram of system at x displacementfrom mean position.

At mean position,

m L kω2 = ∆At x displacement,

ma k x m L x= + − +( ) ( )∆ ω2 = −( )k m xω2

So, ω ωn

k

m= − 2

TEST Edge Question can be asked by changing thecourse of motion of the system. We can usecombination of springs also. We should be sureabout the centripetal force which is providing thecirculatory motion.

7. (b) Idea We have to calculate the image formed bythe concave lens. The image so formed willacts as object for the convex lens. Then, we willapply the lens formula

1 1 1

v u f− =

We can also use the formula for focal length ofcombination of lens

1 1 1

1 2f f feq

= + +...

Image formation from1st lens1 1 1v u f

− =

1 180

140v

+ =−

⇒ v = −803

cm

For second lens,1 1 1v u f

− =

1 180 3

140v

−−

=/

v = −80 cm

TEST Edge Different questions can be formed bychanging the configuration of the lens system.

i.e., we can use convex-concave, one lens can also bepolished

The points should be kept in mind

(i) Image formed by first lens or mirror will beobject for the second lens or mirror

(ii) Paraxial rays will be considered.

PRACTICE SET 5 285

m

x∆

LNL Mean position

m L xω2 ( + )

mg

N

v0

8. (c) Idea To solve these types of questions we have toapply equation for energy of photon

i.e., E hvhc= =λ

λmin = hc

eV

Wavelength λk is independent of the acceleratingvoltage ( )V , while the minimum wavelength λc isinversely proportional to v. Therefore as v isincreased, λk remains unchanged whereas λc

decreases or λ λk c− will increases.

TEST Edge Questions could be framed by varying thepotential of the used tube. If potential changes thenaccordingly λ will vary and different types ofquestions can be formed.

(i) We should be aware about the wavelength andthe potential applied.

(ii) According to the frequency observed we willdecide the type like K K Kα β γ...

1 1

9. (c) Idea It is based on standing waves on a stringfixed at both ends and to calculate velocity,fundamental wavelength of string.

i.e., fL

F= πµ2

where f is frequency of standing wave and µ ismass per unit length.

Free body diagram at angle θ (small)

So, restoring torque about O

τ θ θ θR ka ka mgL= + −2 2

= −( )2 2ka mgL θFor unstable system

2 02ka mgL− ≤because then torque will not be restoring

So, 2 2ka mgL=L

a

k

mg22=

TEST Edge This idea is important according to JEEAdvanced point of view. Students should focus onthis concept and relate to laws of vibration ofstretched string.

If a wave enter a region where the wave velocity issmaller, the reflected wave is inverted as shown inFig. (a). If it enters a region where the wave velocityis larger, then the reflected wave is not inverted.

The transmitted wave is never inverted as shown inFig. (b)

10. (b) Idea This problem is based on electrostaticpotential energy i.e., amount of work done inbringing the unit positive charge from onepoint to the other against the electric field.

i.e., V VW

qB A

AB− =0

UQq

a

q

ai = +

×1

42

4 20

2

0πε π ε

UQq

a x

Qq

a x

q

af =

−+

++

×1

41

4 4 20 0

2

0πε πε π ε

=−

+×

14

24 20

2 2

2

0πε π εQqa

a x

q

a

= −

+

×

−2

41

4 20

2

2

1 2

0

qQ

a

x

a

q

aπε π ε

∴ ∆U U UqQ

axf i= − =

24 0

2

πε

TEST Edge This concept is important according toJEE Advanced. In every year, one question is askedfrom this topic. So, student should concentrate onthis idea and relate to electric dipole, torque andpotential energy of a dipole placed in a uniformelectric field.

Just as the electric field due to a collection of pointcharges is the vector sum of the fields produced byeach charge, the electric potential due to a collectionof point charge is the scalar sum of the potential dueto each charge.

i.e., Vq

ri

ii

= ∑1

4 0πε

11. (a,c,d) Idea As we know, this problem is based onstanding wave on a string. We have tocalculate frequency, wavelength andvelocity of propagation of transverse wavealong the string end.

General equations for standing wave on string are

fn

l

T=2 µ

⇒ L n= λ2

where, n = mode of vibration,


kaθmg

O

kaθ

(a)

(b)

For fundamental mode n =1

L = ×12λ

,

Now that is clear that λ depend on length that’s why λwill not change with change in tension.

fl

L

T=2 µ

i.e., f T∝

T T T120

100= +

f

f

T

T

T

T1 1

1 21 20100 1

15

= =+

= +

/

f

f1 =1.0954

Q f = =150.0954

157.23 Hz

Q Velocity of propagationv

v

T

T

T

T1 1 1 1 5

115

11

10= = + = + = +( / )

So, 10% increment in velocity.

TEST Edge This idea is important according to JEEAdvanced. In every year, this concept is asked andstudents should relate with stationary waves in aircolumn (i.e., open and closed organ pipes) andDoppler’s effect.

Note We know that ν ν1 02= is the first overtone,

ν ν2 03= is the second overtone. Thus, for a stringfixed at both the ends, all the overtones areharmonics of the fundamental frequency and all theharmonics of the fundamental frequency areovertones. Above property is unique to the stringand make it so valuable in musical instruments suchas violin, sitar and sarod.

12. (a,b,d) Idea This problem is based on wave motion,intensity and loudness of a wave.

i.e., LI

I= 10 10

0

log

where, I0 is the intensity of minimumaudible sound which is 10 12 2− −Wm .

For point source, sound will spread in the sphericalregion. So, intensity at any point A is

IP

r= 0

2 24πWm

So, Ir

∝ 12 .

Loudness of sound in decible will be

LI

I( ) logdB =

10 10

0

[where, I0 = reference intensity]

60 100

=

log

I

I⇒ I I= ×0

610

I = × =− −10 10 1012 6 6 W/m2 [I01210= − W/m2]

P

rI

4102

6

π= = −

rP2

6

3

64 101 10

4 10250=

×= ×

× ×=−

−

−π π3.14

r = 250π

m,

I Ar

∝ ∝2 21

, so Ar

∝ 1

At a given time phase difference between 2 points.

∆ ∆θ πλ

= 2x λ =

v

f

λ = =340170

2 m

∆θ π π= × − =22

55 4 51( )

∆θ π π= +25 2( )

i e. ., when both points will oscillates in opposite phase.

Ar

∝ 1so

A

A

r

r1

2

2

1= ,

2 5542

A

A= , A

A2

855

=

∴ Due to opposite phase that would be written as

− 855A

TEST Edge In every year atleast one question isasked in JEE Advanced. IIT-JEE students shouldconcentrate on this idea and interrelate with soundwaves, speed of sound in a gas, and intensity ofsound waves.

Note The intensity of sound waves is given by

Ip

v= max

2

2ρ

where, pmax is the maximum change of pressure inthe medium.

The intensity of waves emitting in all directions dueto point source

i.e., Ir

∝ 12

13. (a,b,c,d) Idea In solving these types of questions, wehave to find the electric fields individuallydue to both parts then we will calculate thenet field by principle of superposition.

i.e., E E E1 2net = + +...

PRACTICE SET 5 287

P0 Ar

By the principle of superposition of fieldsE E E1 2= +

where, E = net fieldE1 = field due to remaining massE2 = field due to mass in hole = v

E10

30

30

22 2= =

=

=E

GM

ar

GM

a

a GM

a

TEST Edge Questions can be asked by forming holeof different shape. The variety in question can alsobe improved by taking system of varying density.

Consider the situation where in a solid sphere ofradius R a hole is made at position r as shown.

That the field will be as follows

E rinside ∝ ⇒ Er

outside ∝ 12

14. (a,b,c,d) Idea It is based on the concept of elasticcollision, of a body and to calculateaverage force and impulse imparted byparticle.

Rv

gv Rg vmax , ,= = = × =

22 160 10 40 m / s

(a) I P mv= = = × × =−∆ ( ) ( )45 10 403 1.8 Ns

(b) v a x av

x

22 2

222

402 2 10

= = =× × −av av,( ) = 40000 2m/s

F m aav av N= = × × =−45 10 40000 18003

(c) ∆tI

F= = =

av

1.81800

0.001 s

TEST Edge As we know, when two bodies collide,they exert force on each other while in contact suchthat large forces acting for a very short duration arecalled impulsive forces.

The change in momentum produced by such animpulsive force is

p p pp

Ff i

P

P

t

t

t

t

dd

dtdt dt

i

f

i

f

i

f

− = = =∫ ∫ ∫

This quantity Fdtt

t

i

f

∫ is known as the impulse of the

force F during the time interval t i to t f .

15. (a,b,d) Idea When a wire is carrying a current therewill be magnetic field surrounding thewire. The value of magnetic field due to ainfinte length wire of a distance r will be

BI

r=

µπ0

2

Magnetic field at P is B, perpendicular to OP in thedirection shown below.

B i j= −B Bsin $ cos $θ θ

BI

r

y

r

x

r= = =µ

πθ θ0

2, sin , cos

B i ji j= − = −+

µπ

µπ

02

02 22

12

I

ry x

I y x

x y( $ $)

( $ $)( )

r x y2 2 2= +

TEST Edge Questions can be asked by varying theshape of the wire. Suppose we have the shape asshown below

So, the electric field due to the parts AB and AC willbe zero and we will calculate the net field due to thepart BC which will be outward.

16. (4) Idea This problem is based on conservation ofmomentum and frequency in simple harmonic

motion. i.e., Frequency( )ν = 1

2πk

m

where k is the force constant and m is the massof the particle.

The total momentum of the system in the horizontaldirection is conserved. We draw the FBD, assumingthe displacement of the block to be x1 and x2 inopposite directions, and the total extension x is givenby

x x x= +1 2

and m x m x1 1 2 2=

∴ md x

dtk x x1

21

2 1 2= − +( )


RO

r

ti tf

Fo

rce

()

F

Time taken

Impulsive

force

r

θ

B

P x, y( )

x

y

θ

θ O

rB

C

A

DB

m1 m2

x2x1

kx kx

and md x

dtk x x2

22

2 1 2= − +( )

After suitably manipulating the equations, we get

d x

dt

k m m

m mx

22

21 2

1 2= − − + ⋅( )

i e. ., frequency = +12

1 2

1 2πk m m

m m

( )

= +×

12

300 2 32 3( )

~− 2.5 Hz.

∴ Number of complete oscillations in 1 min

= =6024

2.5.

As given 24 6= n

n = 4

TEST Edge Above idea is important according to JEEAdvanced. Every year, two or three questions areasked in this concept and relate with torque, energyconservation and dumped harmonic motion.

Note When the system oscillates with almost thenatural angular frequency k m/ (with which thesystem will oscillate if there is no damping) andwith amplitude decreasing with time according tothe equation

i.e., A A eo

bt

m=−2

Therefore, the amplitude decreases with time andfinally becomes zero as shown in figure.

17. (3) Idea It is based on the thermal conductivity ofthe material and bulk modulus of elasticity.We can directly relate change in length ofsteel bar with stress applied in bar andeasily calculate the thermal coefficient ofexpansion ( )α .

Considering an element of the bar of the length dx

shown in figure

Change in length of element length dx due totemperature difference is

du dx T= × ×α ∆

or du dxT x

L= × ×α 0

2

2

Integrating, we get

∫ ∫=duT x

Ldx

L

0

02

2α

∆ = αT L

L

03

23

Bar is rigidly field

Stress in bar σ = × = ×EL

Estrain∆

or σ α α= × =T L

L LE

E T03

20

3 3.

Stress in bar σ α= E T0

3

As given in question

⇒ E T

x

E Tα α0 0

3=

So, x = 3

TEST Edge In thermal conduction, it is found that insteady state the heat current is proportional to thearea of cross-section A. Proportional to the changein temperature ( )T T1 2− .

Then,∆∆

Q

t

K A T T

x=

−( )1 2

where, K is a constant for material of the slab and iscalled thermal conductivity of the material.

18. (2) Idea This problem is based on Moseley’s law inorder to calculate atomic number of element.

i.e., ∆E h Rhc Z bn n

= = − −

ν ( )2

12

22

1 1

where ∆ E is the change in energy (state).

Kα line wavelength remains same irrespective of theaccelerating voltage

Now,1

111

12

34

122 2

2

λαK

R Z R Z= − −

= −( ) ( ) …(i)

and λ cut offat kV

Å Å( )V V=

= =×10

312400 12400

10 10

=1.248 Å = λmin.1

and λ λcut off

at kV

Å 0.62Å( )

min.

V ==

×= =

103

11240020 10 2

Given,λ λλ λ

α

α

K

K

−−

=min.

min.

2

13

⇒ λ λ λ λα αK K− = −min. min.2 13 3

⇒ λ λαK = =5

4 1min. 1.55 Å …(ii)

Using, Eq. (ii) in (i), we get

ZRK

= + = + ×14

31

43λ

α.

911.51.55

⇒ Z ≈ 29

PRACTICE SET 5 289

dxx

Am

plit

ud

e

Time taken

TEST Edge According to JEE Advanced, this conceptis important and it is generally asked in exams. So,students should focus on Moseley’s law and relatewith Bragg’s law and properties of X-rays.

Note Moseley’s observation can be mathematicallyexpressed as

ν = −a Z b( )

where a and b are constants. This relation is known

as Moseley’s law.

In figure, show a graph ν of X-rays against the

atomic number.

19. (0006) Idea As, we know whenever a ray of lightgoes from a denser medium to a rarermedium, it bends away from the normali.e., based on total internal reflection oflight.

Ray enters at A just reflects at B in TIR.

So, θc n= −sin ( / )1 1

Q Snell’s law at A

sin sinθ =n r

r c+ = °θ 90

r c= ° −90 θsin sin ( ) cosr c c= ° − =90 θ θ

If rays does not come out from second surface thanincidence angle at curved surface should be ≥ θc

So, r should be ≤ ° −90 θc

sin sin ( )r c≤ ° −90 θ

sin sinrn

c≤ − ≤ −1 1122θ

From Snell’s law at A

sinsin

rn

= θ

sin θ ≤ − ≤ −n2 1 1.25 1

sin θ ≤ 12

θ ≤ °30 , θmax = °30So, θ/5 0006⇒

TEST Edge In every year, one or two are asked inJEE Advanced. Student should focus on this ideaand related to Lens maker’s formula and itsmagnification.

For total internal reflection of light take place, thenfollowing conditions must be obeyed.

(i) The ray must travel from denser to rarermedium.

(ii) Angle of incidence ( )Li must be greater than

critical angle ( )Lc i.e., c =

−sin 1 µµ

rarer

denser

where, c is a critical angle.

20. (1) Idea It is based on the conversation of totalmomentum in perfectly elastic collision and tocalculate the pressure exerted by beam on thewall.

Let say mass of each molecule is m, momentumchange for each molecules

= 2mv cos θTotal number of molecules colliding in unit time orpassing through cross-section of beam

= ′n v A( ) ( )

[ ′ =A Area of cross-section of beam]

′ =A A cos θ =n v A( ) ( cos )θSo, total momentum exchange per unit line per unitarea.

∆∆p

tmv nA= 2 2 2cos θ

p F A mv n= =/ cos2 2 2 θ= ×1.0043 N/m2105

TEST Edge According to the principle ofconservation of linear momentum, if no externalforce is acting on a system, then momentum of thesystem remains constant.

⇒ If no force is acting, then F = 0

i.e.,d

dt

p= 0

∴ p = constant

or m v m v1 1 2 2= = constant

21. (a) Idea This problem is based on the concept ofnature of oxide of 3rd period elements. To solvethis problem students are suggested to analysethe reactivity of aluminium oxide with acid orbases.


θ

θ

θ

C

A θc θc

B

r

Atomic number

νin

10

Hz

91

/2

Nature of oxides

A set of quantum numbers

(a) denotes 3 1p which is valence shell electronicconfiguration of Al.

(b) denotes 3 3p which is valence shell electronicconfiguration of P.

(c) denotes 3 2p which is valence shell electronicconfiguration of Si.

(d) denotes 3 1s which is valence shell electronicconfiguration of Na.

The elements of period 3 forms following respectiveoxides as

Element Oxides

Na Na O2 (Alkaline)

Mg MgO (Alkaline)

Al Al O2 3(Amphoteric)

Si SiO2 (Acidic)

P P O P O2 5 4 10≡≡ (Acidic)

S SO ,SO3 2 (Acidic)

Cl Cl O2 7 (Acidic)

out of above oxides the Al O2 3 is the only one whichproduces amphoteric solution after hydrolysis.

TEST Edge Similar problems based on the concept ofnature and reaction of various oxides of metal andnon-metals can also be asked in JEE Advanced.Here, the generalisation is

Metal oxide (ionic) → Alkaline

Non-metal oxide (covalent) → Acidic.

Semi metal oxide (polar covalent) → AmphotericHowever, precisely we can say alkaline ∝ ioniccharacter in oxide. Acidic nature of oxide ∝ covalentcharacter oxide.

22. (d) Idea This problem is based on concept ofdetermination of molecular structure oforganic compound and reaction of phenolwith NaOH. To solve this problem studentsare advised to search the most polar bondwhich is to be broken. It is also advised to becareful as the Ph-Cl bond has double bondcharacter. Hence it will not undergonucleophilic substitution reaction.

While solving this problem students areadvised to undergo following sequentialsteps.

Determine molecular structure of organiccompound using degree of unsaturation.

Then use the concept of resonance to answerthe question asked.

Determination of molecular structure

Degree of unsaturation is 4 and it is an aromaticcompound. So possible structure of comopund is

Alkyl halide are extremely less reactive towards SN

reaction due to following reason.

Resonance effect

C H Cl NaOH6 5 + → No reaction

TEST Edge JEE Advanced, includes these types ofquestions to judge the basic knowledge of studentsregarding nucleophilic substitution reaction.Similar questions based on basic concept ofelectrophilic addition reaction, nucleophilicaddition reaction etc, can also be asked such as

Which type of intermediate is obtained when bromine addof propene?

After undergoing proper mechanism of reactionone can get ‘non-classical carbocation’ as answer.

Note Non-classical and classical carbocation Thecyclic and acyclic carbocation are two types ofcarbocation which are known as non-classical andclassical carbocation respectively.

23. (a) Idea This question is based on conceptualmixing of mole concept, neutralisationreaction and stoichiometry, while solving thistype of problem students are advised to gothrough making road map of given problemand then use the concept of stoichiometry andmole concept to solve this problem.

As 6 mole of I2 is formed during the reactionhence,

x × = ×660

10000.1

x =×

= −66 1000

10 3 mole

TEST Edge Similar problems based on conceptualmixing of mole concept, stoichiometry and titrationreaction can also be asked such as

PRACTICE SET 5 291

Cl Cl

s

Double bond

character

⊕

Cl

Brr

r

Non-classical

carbocation

[bridged carbocation]

classical

carbocation

Generally hard water contains Ca 2+ and Mg 2+ , if 1 L ofpond water contains 20 mg of Ca 2+ and 12 mg of Mg 2+

ions. What is the volume of 2N, Na CO2 3 solutionrequired to soften 5000 L of pond water?

After solving problem one can get answer as 5 litre(use the concept of pH and streochemistry)

24. (a) Idea This problem is based on the concept ofmolecular orbital electronic configuration ofoxides of nitrogen. While solving thisproblem students are advised to keep in mindthe presence of unpaired electron in givenoxide, if molecule/compound containunpaired electron then excitation becomespossible.

Cause of color of NO2

Molecular orbital electronic configuration of NO2confirms the presence of unpaired electron whichcan undergo easy transition from ground stateenergy level to excited state level by absorbing lightof suitable wavelength.

Cause of color of N O2 3

Due to low difference between occupied andunoccupied energy levels of electrons of moleculesof N O2 3 it absorbs a part of visible light spectrumwhich causes color of N O2 3.

TEST Edge Similar problems based on colourmagnetic properties, dimerisation reaction,disproportionation reaction and MOEC (MolecularOrbital Electronic Configuration) can also be askedin JEE Advanced.

25. (d) Idea This problem is based on the concept ofmolar specific heat capacity and Boyle’s law.While solving the problem students areadvised to go through calculation of totalmolar heat capacity first followed by use ofBoyle’s law to calculate final temperature afterexpansion and then calculate total energychanges during expansion.

CR R

RV =× + ×

+=

1 3 232

1 22

TV T Vr r1 1

12 2

1− −=

⇒ T2 160= K

∆E nC T TV= −( )2 1

= × − = −3 2 160 320 960R R( )

TEST Edge In JEE Advanced these types ofquestions are asked to judge the knowledge ofstudents regarding states of matter.

Similar problems with conceptual mixing ofgaseous law, ideal gas equation and work done canalso be calculated. Thus, it is advisable to gothrough connective as well as comparative study ofthese topics.

Remember

For adiabatic process change in internal energy canbe calculated as

∆ E nC T TV= − −( )2 1

= −−

( )

( )

p V p V1 1 2 2

1γ

where, γ =C

CP

V

Value of γ varies for different types of gases

Nature of gases Value of γ

Monoatomic gases 1.66

Diatomic gases 1.40

Polyatomic gases 1.33

26. (c) Idea This problem is based on free radicaladdition reaction to the alkene. To solve thisproblem students are advised to undergoproper mechanism including free radical as anintermediate. The stability of free radicalsgenerated during the reaction play key rolehere.

C H C

O

OOC

O

C H6 5 6 5Benzoyl peroxide

Homolysis

→

2 2C H C

O

O C H6 5CO

6 52

→•

•

↓

Cl C

Cl

Cl

Br

• + CCl C H Br3 6 5

CH CH CH CH CCl3 2 3 == + →•2

CH CH CH CH CCl3 2 2 3

secondary radical

•

↓

Cl C

Cl

Br

Cl

CH CH CH

Br

CH CCl3 2 2 3

TEST Edge Similar problems based on conceptualmixing of nucleophilic addition reaction to α β,unsaturated carbonyl compound and function ofGrignard and Gilmann reagent can also be asked inJEE Advanced, such as


What will be the respective products X and Y when 5methyl cyclohex-2-eneone is treated with Grignardreagent (RMgX) and Gilmann reagents respectively?

After using the concept of nucleophilic additionreaction and HSAB theory, one can get answer as

HSAB theory According to HSAB (Hard and SoftAcid Base) theory hard acid adds to hard base andsoft acid adds to soft base leading to the formationof stable compound.

Here concept of hard electrophile E+ with morepositive charge density soft electrophile, (E+ lesspositive charge density) hard nucleophile and softnucleophile is used.

Grignard reagent Rs

Mgs

x↑

−−

+

H. Nucleophilic

Gilmann reagent R2

↑−

S. Nucleophilic

Cu Li

27. (a) Idea This problem includes conceptual mixingof brown ring test and determination ofmagnetic moment of coordination compound.While solving the problem students areadvised to calculate number of unpairedelectron in complex followed by magneticmoment.

Brown ring test When a freshly prepared FeSO4solution is added to aqueous solution of NO3

− ionfollowed by addition of concentrated H SO2 4 thebrown ring is observed at junction between twoliquids. This colour is due to charge transfer oxidationstate of iron in this complex is + I.

NO NO→ ++ eΘ

Fe Fe2+ + → +eΘ

Magnetic moment can be calculated as

µ = +n n( )2 BM = +3 3 2( ) BM

= 15 BM = 3.87BM

TEST Edge Similar problems based on the conceptof complex formation during qualitative analysis ofinorganic ions (atoms and anions) and theircharacteristics can also be asked in JEE Advanced,

so students are advised to go through study ofqualitative analysis of inorganic compound andcharacteristics of complex.

28. (d) Idea This problem is based on the concept offormation of carbocation, structure ofcarbocation and stability of carbocation.Complete the reaction given in options firstfollowed by looking at the stability andstructure of carbocation.

This is a non-planar carbocation as the cation isdelocalized to all the three rings so this is stable too.The non-planar structure is due to steric heindranceof these three phenyl rings. All three phenyl ring istilted to plane by 45 like wings of a fan.

TEST Edge Similar problems based on conceptualmixing of stability, structure and generation ofcarbon free radical and carbanion can also be asked,such as.

Which of the following produced carbanion hasmaximum stability?

(c) After using the concept of electron withdrawingability anyone can answer this question.

PRACTICE SET 5 293

n = 3n = unpaired electron

O

Hard electrophilic centre

Soft electrophilic centre

C + Ag+

Cl

r

C

45° tilted to

each other

F

SbF5

r

s[SbF ]6

Tropylium

carbocation

(a)

Brr

+ Br+(b)

+ H+r

H

(c)

(d)

CH3

O

X = ,

OH CH3

CH3

O

CN

O

O

NaOEt

LiPh

LiPh

LiPh

NO2

CN

(a)

(b)

(c)

(d)


29. (b) Idea This problem is based on nomenclatureand isomerism of coordination compound.Write the structure of coordination compounddraw their mirror image so and then analyseeither then are non-superimposable orsuperimposable, if they are superimposablethen optically inactive otherwise opticallyactive.

(a) trioxalato chromate (III) [Cr(C O ) ]2 4 33−

(b) trans dichlorobis ethylene diamine platnium (II)chloride

(c) trans diamine dichlorobis ethylene diaminechromium (I)

b and c both are optically inactive,

TEST Edge Similar problems based on concept ofisomerism and optical active including magneticmoment and isomerism in coordination compoundcan also be asked in JEE Advanced, so students are

advised to go through study of these topics.

30. (b) Idea This problem is based on nomenclature oforganic compound. While solving theproblem students are advised to choose themost commonly used term in derived name oforganic compound.

So, name is iso-butyl trimethyl methane.

TEST Edge Similar problems based onnomenclature and isomerism of organic compoundcan also be asked. Remember the priority chart of

functional group to tackle such problems.

31. (a,b,d) Idea This problem is based on conceptualmixing of kinetics of gaseous reaction andhalf-life time of reaction. Calculate partialpressure of substances in term of initialpressure of reactant taken first and thent99%.

3 2 2A B C→ +t = 0; p0 0 0

t = 20; p x0 − 23x 2

3x

t = ∞ 23

0p 23

0p

⇒ 43

40p = ⇒ t50 20% = min is the half-life.

p0 3= atm t75 2 20 40% = × = min

px

0 3+ = 3.5 t t87.5% = ×3 50%

= × =3 20 60 min

x =1.5 t t9923% = × 99.9%

= × ×23

10 50t %

= 4003

TEST Edge Similar problem based on concept ofcollision theory, absolute rate law and Arrheniusequation can also be asked.

32. (a,c,d) Idea This problem includes conceptualmixing of Hall-Heroult process, formationof slag and reduction of zinc. Read all thestatements very carefully then to answerthe question.

Formation of slag When ore is treated with silica itform slag of iron silicates

FeO SiO Fe SiO2 3Slag

+ → ⋅

(b) Reduction of Zn On reaction of ZnO with coke itconverts ZnO into Zn and CO. This is due to reductionof ZnO into Zn.

ZnO C Zn CO+ → +

(c) 2Al O 3C 4Al 3CO2 3 2+ → +

This equation represents electrolysis of aluminiumwhich is done by using electrolytic cell which containsteel cathode and graphite anode.

(d) Monds process is used in refining of nickel whichinvolves conversion of Ni to Ni(CO)4 and thenNi(CO)4is subjected to higher temperature. So that itdecomposes to give pure metals.

CrC O2 4 C O2 4

C O2 4 C O2 4

C O2 4 C O2 4

3– 3–

Mirror Non superimposable

Cr

Pt Pt

Cl Cl

Cl Cl

en en enen

Superimposable mirror images

Cr Cr

Cl

Cl Cl

Cl

en en

Superimposable mirror images

NH3 H N3

H N3NH3

CH —CH—CH —C—CH2 2 3

CH3 CH3

CH3

Methane

unit

Iso octane

TEST Edge Similar problems based on the conceptof purification, reduction and concentration of orecan also be asked in JEE Advanced, some care mustbe taken when roasting and calcination are studiedtogether.

Remember

Roasting is heating of metal ores in presence of airbelow its melting point and calcination is heating ofmetallic ores in absence of air below its meltingpoint.

33. (a) Idea This problem is based on the concept ofelectrophilic addition reaction, free radicalsubstitution reaction, Hoffman bromamidereaction including stereochemistry of theproducts.

TEST Edge Similar problems based on Aldolcondensation, cross-aldol reaction, Cannizzaroreaction, cross-cannizzaro reaction etc., includingstereochemistry of products can also be asked.

What will the correct stereochemistry of product obtainedby chemical reaction between sodium acetate and methylbutanone?

After undergoing proper mechanism you will get

34. (a,b,d) Idea This problem involves conceptualmixing of the structure of diamonds andcorrundum, uses of carbogen and natureof SnCl2.

Carbogen is a mixture of 0-5% CO2 and O2 which isused in artificial respiration of pneumonia patients.

SnCl2 is a powerful reducing agent

PbO is known as litharge which has yellow orangecolour.

TEST Edge Similar problems based on conceptualmixing of preparations properties and uses ofinorganic compounds can be asked in JEEAdvanced, so students are advised to go through indepth study of preparation properties and uses ofPH3, O3, SO3, FeSO4 etc.

35. (a) Idea This problem is based on the concept offunction of reagents and stereochemistry. Theconversion using, given reagents keeping inmind the stereochemistry of the product.

PRACTICE SET 5 295

H⊕ H O2

OH(+)–

⊕H O2

C—NH2Br + KOH2

NH2

Achiral

O

C—H

O

+ HCN

C—CN

OH

H

(+)–

(ii)

(iii)

(iv)

Br2

Br(+)–

UV(i) (Major)

O OH H

as a product.

HH

Structure of diamond

NO2

SnCl2

NH2

Me

C==CH

Me

H

H /H O⊕2

hydrationCH —CH—CH —CH3 2 3

OH

H⊕

∆–H O2

CH —CH—CH —CH3 2 3

⊕ +H⊕ H

C==CMe

Me

H(Major trans)

(a)

Me

C==CH

Me

H

NiCH —CH—CH —CH3 2 3

OH

Br /hv2

CH —CH—CH —CH —CH3 2 2 3

Alc KOHH

C==CMe

Me

H(Major)

(b)

∆

Me

C==CH

Me

H

Br /CH2 4CH —CH—CH—CH3 3

Br

2NaNH2CH —CH C—CH3 3≡≡

liq. NH3+ H

C==CMe

Me

H(Trans)

(c)

Br

∆ Na

Me

C==CH

CH3

H

Br2CH —CH—CH—CH3 3

Br

NAI + Acetene (dehaloge

nating agent for dihalide)H

C==CMe

Me

H(Major)

(d)

Br

CCl4

TEST Edge Similar problems, based onstereochemistry of product and addition reaction tocarbonyl compounds such as aldehyde, ketone,carboxylic acid and derivatives of carboxylic acidscan also be asked in JEE Advanced, related such as

Which of the following is correct regardingstereochemistry of product?

(a) P exists in enantiomeric form

(b) P exists in diastereomeric form

(c) P exists in racemic form

(d) None of the above

After undergoing proper mechanism to solve thesequestion you will get (b) as a answer.

Remember

When there is presence of stereogenic centre at αcarbon of carbonyl compound and it undergoaddition reaction it will produce a diastereomeralways.

36. (4) Idea This problem is based on concept ofdetermination of equilibrium constant atdifferent volume of reaction.

A B C+ 2 s

at V 3M 4M x M

at 2V32

42

22

+

+

−

y y

xy

kx x

eq =×

=3 4 4282

Given 2 2 3y + =

y = =12

0.5

k

x

xeq (1.5 0.5)

=

−

+=

12

3 482

x = 4 M

TEST Edge Similar problem based on the concept ofequilibrium constant solubility product can also beasked. Go through in depth study of the relationbetween equilibrium constant and solubilityproduct.

Remember

If Q = Ionic product and

ksp = Solubility product

Then,

Q ksp> , then precipitation takes place

Q ksp< , then no precipitation takes place

Q ksp= , then reaction is at equilibrium

37. (b) Idea This problem can be solved by using theconcept of electronic transition in Brackettseries and rydberg equation. While solving theproblem students are advised to identify thetype of series in which lines of spectrum isobtained. Use the energy boundary of eachline to choose the correct choice.

∆E = ×34

0.85 eV

Photon will be in Brackett series

(Q 0.31 0.85≤ ≤E ) for Brackett

0.85 13.6114

14

12 2−

= −

n

0.8513.616

114

14 2

−

= −

n

⇒ 4 12n

=

⇒ n = 8

Hence, n = 8

TEST Edge Similar problems based on conceptualmixing of wavelength of light and energy of aparticular transition can also be asked in JEEAdvanced.

Extra care must be taken during calculation ofenergy levels regarding unit of energy conversion.e g. . ,

The ratio of the wave numbers for highest energytransition of electron in Lyman and Balmer series of atomis

Ans. 4 : 1 Calculate through the given formulae

38. (3) Idea This problem is based on oxidation state ofconversion of KMnO4 to MnO2 while solvingthis problem students are advised to write theexact chemical reaction of giventransformation followed by calculating thedifference in oxidation state.

Extra care must be taken regarding medium ofreaction.

The transformation of KMnO4 to MnO2 occurs inacidic medium as follows.

4KMnO 4H 4MnO 3O 2H4 2 2 2+ → + ++ O

x − =4 0

x = + 4

x − + =8 1 0

x = 7

Change in oxidation number = − =7 4 3

TEST Edge Similar problems and regardingtransformation of KMnO4 and K Cr O2 2 7 in differentmedium of reaction can also be asked in JEEAdvanced. Even sometimes equaivalent massquestions can be asked also.


O

CH3NaOEt

CH —CHO3Br

( )P

39. (4) Idea Also this problem is based on theconceptual mixing of molecular formula ofZiegler natta catalyst and oxidation state oftransition element present in the complex.

Zieglar Natta catalyst

⇓ is a mixture of

(C H ) Al TiCl2 5 3 4+ (Thus, oxidation state of Ti is +4)

TEST Edge Similar problems based on redoxreaction or oxidation and reduction reaction ofinorganic and organic compound can also be asked.Such as

What will be the change in oxidation state whenhypochlorous acid reacts with aqueous solution of acid?

Hints

2 2 2 22HCLO H Cl H O( )2( ) ( ) ( )aq aq e g l+ + → ++ −

= 7

40. (2) Idea This problem is based on concept ofmeasurement of emf during titration of acidand base at different end points. While solvingthe problems it is suggestive to calculate thepH first, followed by calculation of Ecell.

pH = + =pk pk2 3

210.5

E Ecell cell pH V= + ⋅ = + × =° 0 059 13805 0 059 10 5 2. . . .

TEST Edge Similar problem based on concept ofelectrochemical series and calculation of emf usingNernst equation for different chemical reaction atdifferent concentration can also be asked.

41. (a) Idea Angle between planes a x b y c z d1 1 1 1 0+ + + =and a x b y c z d2 2 2 2 0+ + + = is

cosθ = + +

+ + + +

a a b b c c

a b c a b c

1 2 1 2 1 2

12

12

12

22

22

22

So, Two planes are perpendicular if

a a b b c c1 2 1 2 1 2 0+ + = if a = + +a a a1 2 3$ $ $i j k and

b = + +b b b1 2 3$ $ $i j k then a b× =

$ $ $i j k

a a a

b b b1 2 3

1 2 3

Distance of a point p x y z( , , )1 1 1 from a plane

ax by cz d+ + + = 0 isax by cz

a b c

1 1 1

2 2 2

+ +

+ +

To find equation of plane P, we have the followinginformation available

P passes through the point ( , , )1 2 1−P is perpendicular to planes

P x y z1 2 2 0≡ − + =and P x y z2 2 4≡ − + =The normal to the plane P x y z1 2 2 0≡ − + =lies along the vector 2 2$ $ $i j k− +

The normal to the plane P x y z2 2 4≡ − + =lies along the vector $ $ $i j k− + 2 .

The normal to both the plane P1 and P2 lies along thevector ( $ $ $ ) ($ $ $ )2 2 2i j k i j k− + × − +

The vector along the normal to the plane P is$ $ $i j k

2 2 11 1 2

−−

The equation of the plane P isx y z

x y

− + −−−

= ⇒ + + =1 2 1

2 2 11 1 2

0 1 0

∴ Distance of the point (1, 2, 2) from the plane

P x y≡ + + =1 0 is1 2 1

1 142

2 2+ +

+= =

∴ k 2 2 2=⇒ k = 2

TEST Edge Questions based on distance between

two planes, equation of planes in different form,

distance between two parallel planes and cross

product of vector are asked. To solve such type of

questions students are advised to understand above

mentioned topic such as distance between two

parallel planes ax by cz d+ + + =1 0 and

ax by cz d+ + + =2 0 is given by dd d

a b c= −

+ +1 2

2 2 2.

42. (a) Idea Angle between two straight line in terms oftheir slopes

tan θ = −+

m m

m m1 2

1 21

Area of triangle = × ×1

2base Altitude

Pythagoras theorem, (Hypotenuse)2 = (Base)2+(Altitude)2

Slope of APb

a

b

a= −

−= −0 2

0 2/

Slope of PQ BQb

a

b

a( )

/= −

−= −0

2 02

PRACTICE SET 5 297

B (0, )b

(0, /2)b

Y

P

Q ( /2, 0)a ( , 0)a

AX

30°

O

∴ tan 301

1 2

1 2° = −

+m m

m m

=

− +

+

b

a

b

a

b

a

22

12

2

=+

32 2 2

ba

a b( )

= 13

⇒ 12 3 3

2 2

aba b= +

Now, AB a b2 2 2 9= + =12

93 3

ab =

= 3

⇒ Area of ∆ = 3 sq. units

TEST Edge Question based on basics of straight lines

such as distance formula, slope formula, etc and

geometrical concept such as area of different

figures, intersecting lines, etc. To solve such type of

questions students are advised to understand

concept of straight line such as area of

parallelogram = p p1 2

sinθ, where p1 and p2 are distance

of perpendicular between two pairs of opposite

sides and θ is the angle between adjacent sides.

43. (d) Idea If y x= [ ] if n x n≤ < + 1, then y n= for the

cubic equation ax bx cx d3 2 0+ + + =derivative f x ax bx c′ = + +( ) 3 22 , the roots isgiven by

xb b ac

a= − ± −2 3

3

In the case where b ac2 3 0− ≤ cubic function ismonotonic.

f x x x a x x k( ) [ ]= − + = − +3 33 3

where [ ]a k= ,

′ = −f x x( ) 3 32

= − +3 1 1( ) ( )x x

For three roots f f( ) ( )− <1 1 0

⇒ ( ) ( )k k+ − <2 2 0

⇒ k ∈ −( , )2 2

⇒ [ ] ( , )a ∈ − 2 2

⇒ [ ] , ,a = −1 0 1

a ∈ −[ , )1 2

TEST Edge Questions based on domain and range of

different functions such as polynomial function,

modulus function, etc and concept of application of

derivatives increasing and decreasing function are

asked. To solve such type of question students are

advised to learn definition of different functions

and concept of increasing function such as function

f x( ) is said to be increasing, if

x x1 2< ⇒ f x f x( ) ( )1 2≤ for this f x′ >( ) 0.

44. (a) Idea By definition of definite integral

a

bf x dx b a∫ = −( ) ( ) ( )φ φ

where φ( )x in anti-derivative of f x( ) in [a b, ] Ifuand v are the differentiable function of x then

u v dx u vdxd u

dxv dx dx.

( )( )= −

∫∫ ∫∫

x f x dx x f x x f x dxa a a20

20 0

2∫ ∫′′ = ′ − ′( ) | ( )| ( )

⇒ a f a x f x f x dxa a20 0

2 1′ − − ∫( ) [ | ( )| ( )

⇒ a f a a f a f x dxa20

2′ − − ∫( ) [ ( ) ( ) ]

⇒ a f a a f a f x dxa20

2 2′ − + ∫( ) ( ) ( )

∴ ′′ = ′∫ x f x dx f20

2014 22014 2014( ) ( ) ( )

− + ∫2 2014 2014 20

2014( ) ( ) ( )f f x dx

⇒ ( ) ( ) .20141

20142 2014

12014

212 ⋅ − ⋅ ⋅ +

⇒ 2014 2 2 2014− + ⇒

TEST Edge In JEE Advanced question based ondefinite integral properties and various method ofintegration such as substitution method, etc., areasked. To solve such type of questions students areadvised to learn concept and properties of definiteintegral such as if f x g x( ) ( )≥ then

a

b

a

bf x dx g x dx∫ ∫≥( ) ( ) (where b a> ).

45. (b) Idea We used trigonometric ratios ofsub-multiple of an angle, 1 2 22+ =cos cos /θ θ

Sum of n terms of GP, Sr

rrn

n

= −−

<1

11,| | and

sin( ) cos90 − =θ θto solve this question.

We know1 2 22+ =cos cos /θ θ

∴ + =14

2 82cos cos /π π

⇒ 112

2 82+ = cos /π

⇒ 2 2 4 82+ = cos /π

⇒ 2 2 2 8+ = cos /π


2–2

+ +–

Adding 2 to both sides

2 2 2 2 2 8+ + = + cos /π

⇒ 2 2 2 2 1 8+ + = +( cos / )π

2 2 2 2 2 162+ + = ( cos / )π

2 2 2 4 162+ + = cos /π

2 2 2 2 24+ + = cos /π

Similarly,

2 2 2 22 1+ + ……

+n

ntimes = cos

π

Now, 112

12

12

122 3 2005+ + + + …… +

=−

−= −

112

112

2 1 2

2006

2005( / )

∴ 2 212

452005

sin −

°

2 90 4512

2005

sin ° − °

= ⋅

2

412

2005

cosπ

= 222007cos

π

= x2006

TEST Edge Question based on trigonometric ratios ofcomplementary and supplementary angles,trigonometric ratios of multiples of an angle andsum of n term of AP series are also asked. To solvesuch type of questions students are advised tounderstand concept of trigonometric ratio such as

cotcot cot

cot3

3

3 1

3

2θ θ θθ

= −−

46. (a) Idea If z n= ( ) /1 1 then z i n= ° + 0°(cos sin ) /0 1

Thus, nth root of unity are 1, α , α 2, …, α n−1

Where α π ππ

= = +en

in

i

n

22 2

cos sin

If z x iy= + then| | {Re( )} {Im( )}z z z= +2 2

Now, x x x xn − = − − − …1 0 1 2( ) ( ) ( )α α α( )x n− −α 1

x

xx x x

n

n

−−

= − − …… − −1

1 1 2 1( ) ( ) ( )α α α

[ ]Qα 0 1 0= =if k

⇒ 1 2 11 2+ + + …… + = − −−x x x x xn ( ) ( )α α…… − −( )x nα 1

Putting x =1n n= − − …… − −( ) ( ) ( )1 1 11 2 1α α α

Taking modulus on both sides, we have| | | | ( )1 1 11 2 1− − …… − =−α α α n n

Now | | cos sin1 12 2− = − −α π π

k

k

ni

k

n

= −2 22sinsin cosk

ni

k

n

k

n

π π π

=

−

2 sin

sin cosk

n

k

ni

k

n

π π π

= 2sin k

n

π

∴ | | | | | |1 1 11 2 1− − …… − =−α α α n n

⇒ 2 22

21

sinsin

...sin ( )π π π

n n

n

n

−

= n

22 11n n

n

n

nn− …… − =sin /

sin sin ( )π π π

TEST Edge In JEE Advanced questions based on

properties of modulus of complex numbers,

arguments and properties of nth roots of unity are

asked. To solve such type of questions students are

advised to learn properties of complex number such

as if| | ,| | ,z z1 21 1≤ ≤ then

| | (| | | |) [arg( ) arg( )]z z z z z z1 22

1 22

1 22− ≤ − + − .

47. (b) Idea Apply basic geometrical concepts of parallellines, triangles coplanarity and collinearityeasity of lines with the application ofpermutation and combination i e. . , number oftriangles formed by joining n points out ofwhich m are collinear is n mC C3 3− .

The straight lines L L1 2, and L3 are parallel and lie inthe same plane.

Total number of points are = + +m n k

The total number of triangles formed by these points= + +m n kC3

⇒ but out of these( )m n k+ + points,m points lie onL1n points lie on L2 and k point, lie on L3 and by joiningthese points on the same line we do not get a triangle.

Hence the total number of

∆ = − − −+ +m n k m n kC C C C3 3 3 3

TEST Edge Question based on functionalapplications of permutation and combination areasked. To solve such type of questions students areadvised to learn functional application such as thenumber of all arrangements of n different objectstaken r at time.

(a) When particular object is always indude inarrangement is n

rC r−− ×11 !.

(b) When a particular object is never taken in eacharrangement is n

rC r− ×1 !.

PRACTICE SET 5 299

48. (c) Idea By the triangle law of inequality

| | | | | | | |z z z z z z1 2 3 1 2 3+ + ≤ + +But| | | | | | | |z z z z z z1 2 3 1 2 3+ + = + +When all z z z1 2 3 0, , ≥ or ≤ 0 and trigonometricidentities. cos sin2 21θ θ= − .

Based on property of modulus

Given equation is,

| sin | cos sin | |2 2 2 2 226 2 25x x x x x+ − = + + −

Now, cos sin ( )2 2 22 25x x x+ + −

1 252 2+ + −sin x x

= + −sin2 226x x

i e. ., | | | | | | | |x y z x y z+ + = + +

(where all x y z, , ≥ ≤0 0or )

⇒ 25 02− ≥x

⇒ x 2 25 0− ≤

x ∈ −[ , ]5 5

TEST Edge In JEE Advanced questions based onposition of roots of a quadratic equation, maximumand minimum value of quadratic equation andairthmetic-geometric mean inequality are asked. Tosolve such type of questions students are advised tounderstand concept of airthmetic- geometric meaninequality such as if a b, > 0 and a b≠ , thena b

ab

a b

+ > >+2

21 1

.

49. (b) Idea We applied definition of probability so the

general expression for the probability P of

occurance of an event used in the solution is

P = Measure of the specified part of the region

Measure of the whole region

Where, measure means length or area or

volume of the region, when we are dealing

with one, two or three dimensional space

respectively.

Total number of ways = 643C

The number of ways of selecting squares consistingof 4 unit squares is 7 7 49× =Also each squares with four unit squares form 4L-shapes consisting of 3 squaresNumber of favourable outcomes

= × ×7 7 4

=196

Required probability = 196 643

/ C

TEST Edge Questions based on Poisson distribution,binomial distribution, etc are also asked fromadvance probability. To solve such type ofquestions students are advised to understand theabove mentioned concept such as mode of binomialdistribution is ( )( ) ( )n P r n P+ − ≤ ≤ +1 1 1 .

50. (c) Idea Apply the concept of differentiation i e. . ,

multiplication rule of differentiation, andsolution of differential equation by themethod of variable separable i e. . ,

If general form of equation is

f x dx g y dy( ) ( )+ = 0 then solution is

f x dx g y dy c( ) ( )= +∫∫Consider the given differential equation.

x xdy

dxx y

f xy

f xy( ) ( )

( )( )

+ + + =′

1 1

⇒ x dy

dxy

x

f xy

f xy+ =

+ ′1

1( )( )( )

⇒ x dy y dx

dx x

f xy

f xy

+ =+ ′1

1( )( )

⇒ d xydx

x

f xy

f xy( )

( )( )

=+ ′1

On separating the variables, we getf xy

f xyd xy

dx

x

′ =+

( )( )

( )( )1

Now on intergating both sides, we getf xy

f xyd xy

dx

x

′ =+∫∫

( )( )

( )( )1

⇒ log ( ( )) log ( ) logf xy x c= + +1

⇒ log ( ( )) log ( ( ))f xy c x= +1

[ logQ log = log ( . )]m n m n+⇒ f xy x c( ) ( ) ,= +1 where c being an arbitraryconstant.

TEST Edge In JEE Advance question based onsolution of homogeneous differential equations andlinear differential equation are also asked. To solvesuch type of questions students are advised to learnthe method of solving mentioned differentialequation such as linear equation of the formdx

dyPx Q+ = then it solution is X. IF = +∫ Q dy c.IF

where IF = ∫eP dy

51. (b,c,d) Idea Dot product of two non-zero vectorsinclined at an angle θ is given asa b⋅ = ≤ ≤abcos ,θ θ π0

Scalar triple product of three vectors a, band c is ( | || || |sin cosa b) c a b c× ⋅ = θ φwhere θ is the angle between a and b φ isthe angle between a b× and c. It is alsorepresented as [ ].a b c

Now, u v u v⋅ = | | | | cos θ = cos θGiven, w w u v+ × =⇒ u w w u u v⋅ + × = ⋅( )⇒ u w u v u w u⋅ = ⋅ = =cos [ ]θ 0also v w w u v v⋅ + × = ⋅( )

v w v w u⋅ + =[ ] 1


v w⋅ + =β 1 ( [ ])Qβ = u v w

v w⋅ = −1 βSimilarly, w w w u w v⋅ + × = ⋅( )

w w w v⋅ = ⋅ = −1 β

also β2 2= [ ]u v w =⋅ ⋅ ⋅⋅ ⋅ ⋅⋅ ⋅ ⋅

u u u v u w

v u v v v w

w u w v w w

βθ θ

θ βθ β β

21

1 11 1

= −− −

cos coscoscos

= − −β β θ( cos )1 2

( | | , cos ,Qu u u u v⋅ = = ⋅ =2 1 θw w w v⋅ = ⋅ = −1 β)

β β θ β2 2= −(sin ) ⇒ β θ β= −sin2

⇒ β θ= 12

2sin

TEST Edge Generally, in JEE Advanced crossproduct, vector triple product and addition ofvectors related questions are asked. To solve suchtype of questions students are advised tounderstand the mentioned concept and alsoacquainted yourself with the properties of crossproduct such as

a b b a× ≠ × and ( | | | | ( )a b) a b a b2× = − −2 2 2

52. (b,d) Idea Apply intersecting chords theoremconcept of intersecting chord inside the circleand the concept of airthmetic-geometric meaninequality i e. . , AM GM≥

Q Chord PT QRand intersect each other then

PS ST QS SR× = × …(i)

Now, AM > GM1 1

21 1PS ST

PS ST

+> ×

⇒ 1 1 2PS ST PS ST

+ >×

⇒ 1 1 2PS ST QR SR

+ >×

[Using Eq. (i)] …(ii)

AgainQS SR

QS SR+ > ×2

⇒ 1 2QS SR QS SR×

>+

⇒ 1 2QS SR QR×

> ( )QQS SR QR+ =

⇒ 2 4QS SR QR×

> …(iii)

From Eqs. (ii) and (iii), we have1 1 4

PS ST QR+ >

Hence, answer option (b) and (d) are correct.

TEST Edge Questions based on solution set ofinequality equation, properties of inequality,tangent of a circle, sector of a circle are asked in JEEAdvanced. To solve such type of questions studentsare advised to understand the concept ofinequalities and also acquainted yourself withtheorem of circles such as alternate segmenttheorem and equation of normal of slope m to thecircle x y r2 2 2+ = is my x r m= − ± +1 2 .

53. (a,b,c,d) Idea To solve this problem, use the conceptof algebra of functions, periodic functioni e. . , f X Y: → is a periodic function iff x T f x( ) ( )+ = ∀ ∈x X and properties ofperiodic functions.

f xf x

f x( )

( )( )

+ = −−

153

...(i)

⇒ f x f x f x f x( ) ( ) ( ) ( )+ − − = −1 3 1 5

⇒ f x f x f x( ) [ ( ) ] ( )+ − = + −1 1 3 1 5

⇒ f xf x

f x( )

( )( )

= + −+ −

3 1 51 1

Replacing x by x −1, we get

f xf x

f x( )

( )( )

− = −−

13 5

1…(ii)

Using Eq. (i), f xf x

f x( )

( )( )

+ = + −+ −

21 51 3

=

−−

−

−−

−

f x

f x

f x

f x

( )( )( )( )

53

5

53

3

= −−

2 52

f x

f x

( )( )

…(iii)

Using Eq. (ii),

f xf x

f x

f x

f x

f( )

( )( )

( )( )(

− = − −− −

=

−−

−

23 1 5

1 1

33 5

15

3 x

f x

)( )

−−

−51

1

= −−

2 52

f x

f x

( )( )

…(iv)

From Eqs. (iii) and (iv), we have f x f x( ) ( )+ = −2 2

⇒ f x f x f x( ) ( ) ( )+ = ⇒4 is periodic with period 4

TEST Edge In JEE Advanced domain and range ofreal function, equality of functions, inversefunction and types of functions related questionsare asked. To solve such type of questions studentsare advised to understand above topics and alsoacquainted yourself with the properties ofcomposition of functions such as if f g, and n arefunctions from R to R

then ( )f g oh foh goh+ = +( ) ( ) ( )f g oh foh goh⋅ = ⋅

PRACTICE SET 5 301

P

T

O

SRQ

54. (a, b) Idea To solve this problem, use the conceptof representation of line equation incarterian coordinate system, propertiesand area of rhombus and cosine rule i e. . , inany triangle ABC. Cosing of an angle canexpress in terms of sides

cos ,Ab c a

bc= + −2 2 2

2

cos Bc a b

ac= + −2 2 2

2

and cos Ca b c

ab= + −2 2 2

2

Let the length of the sides of the rhombus be l units.

⇒ Area = °l2 30sin

(since, the angle between consecutive sides is 30°)

⇒ 2 302= °l sin

⇒ l = 2 units

Now, applying cosine rule in ∆ OAB, we get

OB OA AB OA OA2 2 2 2 2 150= + − − °. cos

= + − ⋅ −

4 4 2 4

32

( )

= +4 2 3( )

⇒ OB = +2 2 3

Since, the angle between consecutive sides is 30°,hence the slope of OB is equal to 1.

∴ Coordinates of the point B are given by

B2 2 3

22 2 3

2+ +

,

(using polar coordinates ( cos , sin )r rθ θHere asθ = ° ∠ = °45 45( )BOX

⇒ B = +( ,4 2 3

4 2 3 1 3 1 3+ ≡ + +) ( , )

also since the rhombus can lie entirely in I or IIIquadrants, hence coordinates of B can be

( , )1 3 1 3+ +or ( , )− − − −1 3 1 3

TEST Edge In JEE Advanced questions related todistance formula, section formula and differentpoints related to a triangle such as centroid,incentre, etc are asked. To solve such type ofquestions students are advised to understandfundamentals of carterian coordinate system suchas quadrants, condition of collinearity and alsoacquainted yourself with the properties of trianglessuch as Napier’s analogy i e. . , In any ∆ABC,

tan cotC A c a

c a

B− = −+2 2

55. (a,c) Idea When two sides and angle between themis known, then area of triangle is given by

∆ = 1

2ab Csin , ∆ = 1

2bc Asin

∆ = 1

2ca Bsin

∆ ∆ ∆= +1 212

12 2

ab C b CFC

sin ( ) sin=

+ 12

2( ) ( ) sin ( / )a CF C

⇒ ab C CF C a bsin sin ( / ) ( )= +2

⇒ CFab C

C a b=

+sin

sin / ( )2

=+

2 2 22

ab C C

C a b

sin / cos /sin / ( )

CFab C

a b=

+2 2cos /

also in ∆ CFB

CF

B

a

sin sin ( )=

−π θ(using sine rule)

CF

B

a

sin sin=

θ

CF

B

a

BCsin sin

=+

2

( / )θ = +B C 2

⇒ CFa B

B C=

+sin

sin ( / )2

=+

b A

BC

sin

sin2

usingsinerulea

A

b

Bsin sin=


Y

XO

150°C

A

B

30°

60°

C

A BF

C/2C/2

θ

TEST Edge In JEE Advanced trigonometry i e. . , ratios

of half angle of triangles, different type of circleconnected with triangle and regular polygonrelated questions are asked. To solve such type ofquestions students are advised to understand theapplication of trigonometry and properties oftriangle such as radius of in circle of a triangle isgiven by

ra

B C

A=

sin sin

cos

2 2

2

=b

A C

B

sin sin

cos

2 2

2

=C

B A

C

sin sin

cos

2 2

2

56. (7) Idea For greatest integer function [ ],x if

n x n≤ < + 1, then [ ]x n= and use the concept ofproperty of definite integral i e. . ,

a

b

a

c

c

bf x dx f x dx f x dx∫ ∫ ∫= +( ) ( ) ( ) ( )a c b< <

f x x( ) = +2 1 and α ( ) [ ]x x=From property

f x d x x d f xa

bb( ) ( ) ( ) ( )α α+ =∫∫ 0

α α( ) ( ) ( ) ( )b f b a f a−

⇒ + + = + −∫∫ ( ) [ ] [ ] [ ] ( )x d x x x dx20

2

0

21 2 2 4 1 0

=10

( ) [ ] [ ]x d x x x dx20

2

0

21 10 2+ = − ∫∫

= − +∫∫10 2 0

1

2

0

1dx x dx

= − +

10 2 0

2

2

1

2x

= − −

10 242

12

= − ⋅10 232

10 3 7− =

TEST Edge Questions based on definite integrationby substitution, definite integral as a limit of sumand types of functions i e. . , one-one function, ontofunction, etc are asked in JEE Advanced. To solvesuch type of questions students are advised tounderstand above concept and also acquaintedyourself with approximation in definite integrationsuch as

a

b

a

b

a

bf x g x dx f x dx g x dx∫ ∫ ∫≤( ) ( ) { ( ) }{ ( ) }2 2

57. (2) Idea Apply the concept of area of triangle whenco-ordinates of three points is known i e. . ,

Area of triangle = 1

2

1

1

1

1 1

2 2

3 3

x y

x y

x y

and projection

of points in different quadrants

The coordinates of vertices of projected triangle willbe

′ − ′ − ′A B C( , , ), ( , , ), ( , , )1 1 0 1 1 0 1 1 0

⇒ Area of triangle =−

−12

1 1 11 1 11 1 1

= 2 square units

TEST Edge Question based on direction cosines,direction ratios, angle between two lines,orthocentre of triangle, excentre of triangle, etcrelated to three dimensional geometry are asked inJEE Advanced. To solve such type of questionsstudents are advised to learn the above mentionedconcepts and also acquaint yourself with differentequation of plane such as intercept form i e. . ,x

a

y

b

z

c+ + = 1.

58. (6) Idea Apply the concept of circle touching eachother externally then distance between thecentres equals the sum of their radii i.e.,| |C C r r1 2 1 2= + and the definition of hyperbola tocalculate the eccentricity.

Let us assume that C C1 2and are fixed circlestouching each other externally.

Since C1 touches C2 externally

∴ C C r r1 2 1 2= +

Let us also assume that C is a variable circle whichtouches both of C1 and C2 externally.

Since C touches C1 externally.

∴ CC r r1 1= +

Since, C touches C2 externally.

∴ CC r r2 2= +

∴ | | | |CC CC r r1 2 1 2− = − =Constant

PRACTICE SET 5 303

r1r2

r2

rr

r1

C1

C2

C

Fixed circleFixed circle

Variable circle


∴ Locus of C is a hyperbola, whose eccentricity is

= +−

= +−

| || |

| || |

C C

C C

r r

r r1 2

1 2

1 2

1 2

Now,r

r1

2

75

=

⇒ | || |r r

r r1 2

1 2

7 57 5

+−

= +−

= 6

So, the curve is hyperbola and er r

r r= +

−=1 2

1 26

∴ k = 6

TEST Edge In JEE Advanced question based onparabola and its related terms such as equation ofdirectrix, latus rectum, vertex, etc and ellipse areasked. To solve such type of questions students areadvised to understand various terms related toparabola and ellipse such as pairs of tangent of

ellipsex

a

y

b

2

2

2

2+ is SS T12=

where, Sx

a

y

b= + −

2

2

2

2 1, Sx

a

y

b1

12

212

2 1= + − and

Txx

a

yy

b= + −1

21

2 1

59. (1) Idea Apply the concept of dot producta b⋅ = abcosθ and cross product i e. . ,a b n× = absin $θ and also basic algebraicformula to solve this type of questions.

Now, a b a c b c b c⋅ = ⋅ = ⋅ =0 0 3, , | | | | cos /πb c⋅ =1 2/ [ ]Qa b,a c⊥ ⊥

| | | | | | sina b a b× = °90

= 1

| | | | | | sina c a c× = ° =90 1

Now | | | |a b a c a b× − × = ×2 2

+ × − × ⋅ ×| | ( ) ( )a c a b a c2 2

= + − × × °1 1 2 60| | | | cosa b a c

= −2 2 1 112

( ) ( )

=1

(Q the angle between a b× and a c× = anglebetween b cand )

TEST Edge In JEE Advanced representation ofvector, types of vector such as collinear vector,coplanar vector, etc, addition of vectors relatedquestions are asked. To solved such type ofquestions students are advised to learn the basicfacts of vectors and also acquainted yourself withthe concept of product of three vectors such asscalar triple product i e. . , ( | || || |a b) c a b c× ⋅ =sin cosθ φ and vector triple product i e. . ,a b c a c b a b c× × = ⋅ − ⋅( ) ( ) ( ) .

60. (0) Idea Apply the concept of transformation of onetrigonometric ratio to another trigonometric

ratio i e. . , tansin

cosx

x

x= , etc and concept of

inequality i e. . , if x > 0, xx

+ ≥12..

| sin cos | (tan cot )x x x x+ + =2 3

| sin cos | | tan cot |x x x x+ + = 3

| sin cos |sin cos

x xx x

+ =13

| sin cos || sin cos |

x xx x

+ =13

We know if x xx

> + ≥01

2,

∴ | sin cos || sin cos |

x xx x

+ ≥12

Hence, | sin cos || sin cos |

x xx x

+ =13 has no

solution.

TEST Edge In JEE Advanced questions based onarithmetic-geometric mean inequality i e. . , AM andGM solution of inequalities and trigonometricidentities are asked. To solve such type of questionsstudents are advised to learn the concept of solutionof trigonometric inequalities by graph and alsoacquainted yourself with the concept ofarithmetic-geometric mean inequality such asa a a

na a an

nn1 2

1 21+ +…+ ≥ ⋅ … ≥( ) / n

a a an

1 1 1

1 2

+ +…+

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