VIJETA (JP) IPT-1 Solution Booklet English 22-04-2012[1]

RESONANCE Page No. - 1

QUESTION & SOLUTIONS

DATE : 22-04-2012COURSE NAME : VIJETA (JP)

PAPER-1

PART-I (PHYSICS)

SECTION - IStraight Objective Type

This section contains 30 multiple choice questions. Each question has choices (A), (B), (C) and (D), outof which ONLY ONE is correct.

1. A particle performing SHM takes time equal to T (time period of SHM) in consecutive appearances at a perticularpoint. This point is :(A) An extreme position(B) The mean position(C) Between positive extreme and mean position(D) Between negative extreme and mean position

Ans. (A)Sol. Position where we see the particle once in a time period that is only extreme position. twice through every other

position

2. For a particle performing SHM :(A) The kinetic energy is never equal to the potential energy(B) the kinetic energy is always equal to the potential energy(C) The average kinetic energy in one time period is equal to the average potential energy in this period(D) The average kinetic energy in any time interval is equal to average potential energy in that interval

Ans. (C)

Sol. PAV

= 41

KA2 and KAVV

= 41

KA2

3. A ball of mass m kg hangs from a spring of spring constant k. The ball oscillates with a period of T seconds. Ifthe ball is removed, the spring is shortened(w.r.t. length in mean position) by

(A) 2

2

)2(

Tg

metre (B) 2

2

)2(

gT3

metre (C)

kTm

metre (D) mTk

metre

Ans. (A)

Sol. T = 2 Km

m = k4

T2

2

mg = Kx x = K

mg

x = 2

2

4

KT

Kg

x = 2

2

4

gT

4. A simple pendulum performing SHM has some time period T. What will be the percentage change in its timeperiod if its amplitude is decreased by 5%?(A) 6 % (B) 3 % (C) 1.5 % (D) 0 %

Ans. (D)

ISEET PREPARATORY

PART TEST-1 IPT (PT-1)TARGET : ISEET (IIT-JEE + AIEEE) 2013


Sol. T = 2 g

, As it does not depend on amplitude

% change in time period is 0 % Hence option (D) is correct.

5. The potential energy of a particle of mass 'm' situated in a unidimensional potential field varies as

U(x) =

2

axcos1U0 , where U

0 and a are positive constants. The time period of small oscillations of the

particle about the mean position �

(A) 0

2Ua

m(B) 2

02Ua

m2(C) 2

02Ua

m(D) 4

02Ua

m

Ans. (D)

Sol. Restoring force F = dxdu�

= dx

d�

2ax

cos1U0

F(x) = � u0 2a

sin2ax

for small angle sin 2ax

2ax

F = � u0

4xa2

acc. = m4

xau 20

= �2x =

2

T2

× x

So, Time period T = 40

2Ua

m

6. A simple pendulum ; a physical pendulum; a torsional pendulum and a spring�mass system, each of

same frequency are taken to the Moon. If frequencies are measured on the moon, which system orsystems will have it unchanged ?(A) spring�mass system and torsional pendulum.

(B) only spring�mass system.

(C) spring�mass system and physical pendulum.

(D) physical pendulum and torsional pendulumAns. (A)

Sol. For simple pendulum T = 2 g

changeg

const

For physical pendulum T = 2 mg

(g�change) ,m, const

For Torsional pendulum T = 2 C

Const

C for any planet

For spring mass system T = 2 Km

mg = k

planet all for Const

'ggKm

Both the spring�mass system & torsional pendulum have no dependence on gravitational acceleration for

their time periods.


7. The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period ofoscillation of the bob is t

0 in air. Neglecting frictional force of water and given that the density of the bob is

2 × 1000 kg/m3. Which relationship between t and t0 is true? (Density of water = 1000 kg/m3)

(A) t = t0

(B) t = 0t 2 (C) t = 2t0

(D) 2

t0

Ans. (B)Sol. The time period of simple pendulum in air

T = t0 = 2

g

........... (i)

, being the length of simple pendulum.In water, effective weight of bobw� = weight of bob in air � upthrust

Vgeff

= mg � m�g

=Vg � � Vg = (�� )Vg

where = density of bob,´ = density of water

geff

=

'�g =

'�1 g

t = 2

g

'�1ñ

ñ

(ii)

Thus, 0tt

=

'�1

1

= 2

100021000

1

1

t = 0t 2 .

8. A certain transverse sinusoidal wave of wavelength 20 cm is moving in the positive x direction. The transversevelocity of the particle at x = 0 as a function of time is shown. The amplitude of the motion is :

(A)

5cm (B)

2

cm (C)

10cm (D)

20cm

Ans. (D)

Sol. Vmax

= A = 10 A 4

2 = 10

A =

20cm.


9. A string of length �� is fixed at both ends. It is vibrating in its 3rd overtone with maximum amplitude �a�. The

amplitude at a distance 3

from one end is :

(A) a (B) 0 (C) 2

a3(D)

2a

Ans. (C)

Sol. For third overtone 2

4 = 2 = =

2

As x = 0 is a node

As..

= A sin

2 x

= a sin

31

2

2= a

23

10. The transverse displacement y(x,t) of a wave on a string is given by

xt ab2btax 22

e)t,x(y , where a, b are positive constants x, y and t are standard variable

This represents a :

(A) wave moving in +x direction with speed ba

(B) wave moving in �x direction with speed ab

(C) standing wave of frequency b (D) standing wave of frequency b

1

Ans. (B)

Sol.2]tbxa[e)t,x(y

It is transverse type2)btax(e)t,x(y

Speed v = a

b

and wave is moving along �x direction.

11. A machine gun is mounted on an armored car moving with a speed of 20 ms�1. The gun can point againstthe direction of motion of car. The muzzle speed of bullet is equal to speed of sound in air i.e., 340 ms�1.The time difference between bullet actually reaching and sound of firing reaching at a target 1088 m awayfrom car at the instant of firing is

(A) 1.2 s (B) 0.1 s (C) 1 s (D) 0.2 sAns. (D)

Sol. time to reach sound wave = 340

1088

time to reach bullet = 20�3401088

= 320

1088

t = 1088

3401

�320

1= 1088 ×

34032020

= 0.2 sec


12. Sound signal is sent through a composite tube as shown in the figure. The radius of the quartercircularportion of the tube is r. Speed of sound in air is v. The source of sound is capable of giving varied frequenciesin the range of

1 and

2 (where

2 >

1). If n is an integer then frequency for maximum intensity at point B

is given by :

(A) r

vn

(B) r

vn2

(C) r 2

vn

(D) rvn2

Ans. (B)

Sol. path difference =

nr2r2r2

S = r (2

) n = S

for constructive interference

n = r (2

) = n2r

=

V =

rVn2

13. The fundamental frequency of a closed organ pipe is same as the first overtone frequency of an open pipe.If the length of open pipe is 50 cm, the length of closed pipe is : (speed of sound wave is same in bothorgan pipe)(A) 25 cm (B) 12.5 cm (C) 100 cm (D) 200 cm

Ans. (B)Sol. Closed Open

14V

=2

V

2

= 41

1

=42 =

450

= 12.5 cm

14. A source of sound and an observer both start moving simultaneously from origin, one along X-axis and theother along Y-axis with speed of source equal to twice the speed of observer. The graph between the apparentfrequency (n'

) observed by observer and time t would be : (n is the frequency of the source)

(A) (B) (C) (D)

Ans. (B)

Sol.


n =

cosV�VsinV�V

s

o n tan =

21

is constant and n remains constant and n < n.

so. graph. must be

15. A non�conducting cylinder fitted with a piston contains an ideal monoatomic gas at a temperature of 400 K. The

piston is held fixed while heat Q is given to the gas, It is found the temperature of the gas has increased by 20K. In an isobaric process the same Q heat is supplied slowly to it. What is the change in temperature in thesecond process?(A) 10 K (B) 12 K (C) 20 K (D) 36/5 K

Ans. (B)Sol. Process 1 is isochoric

Q = nCV T

1

Q = n . 23

R T1

Process 2 is isobaric Q = n C

P. T

2

Q = n. 25

R T2

= 2

3nRT

1, 3T

1 = 5T

2 , T

2 = 12 K.

16. If Q amount of heat is given to a diatomic ideal gas in a process in which the gas perform a work 3Q2

on its

surrounding. What is the molar heat capacity (in terms of R) for the process.(A) 2.5 R (B) 5 R (C) 7.5 R (D) 15 R

Ans. (C)Sol. Q = n CT

As Q = nC T C = Tn

Q

...........(i)

Also U = Q � 3Q

3Q2

nCVT =

3Q

, 3Q

2TR5n

nT =

R15Q2

......(ii)

From (i) and (ii)

C = Q2

R15Q = 7.5 R.

17. One mole of an ideal monoatomic gas

35

is mixed with one mole of a diatomic gas

57

. (

denotes the ratio of specific heat at constant pressure, to that at constant volume). What is the value ofadiabatic exponent for the mixture of gases ?

(A) 23

(B) 32

(C) 57

(D) 35

Ans. (A)

Sol. mix = Vmix

Pmix

CC

=R

25

1R23

1

R27

1R25

1

=23

.

18. An ice block at 0°C is dropped from height �h� above the ground. What should be the value of �h� so that it

just melts completely by the time it reaches the bottom assuming the loss of whole gravitational potentialenergy is used as heat by the ice ? [Given : L

f = 80 cal/gm]

(A) 33.6 m (B) 33.6 km (C) 8 m (D) 8 kmAns. (B)


Sol. Applying energy conservation :mgh = mL

f

h = gL f

= 2s/m10

gm/cal80 = 2m/s 10

J/kg 10002.480 =

1010336 3

kgsN 2

= 33.6 km. Ans

19. An ideal gas is initially at temperature T and volume V. Its volume is increased by V due to an increase in

temperature T ,pressure remaining constant. The quantity TV

V

varies with temperature as:

(A) (B) (C) (D)

Ans. (C)Sol. For an ideal gas

PV = nRTP.V = nRT (P = constant)

TV

=

PnR

=

VnRTnR

= TV

T.V

V

=

T1

= T1

or T = 1

graph will be rectangular hyperbola

20. An ideal gas is expanding such that PT3 = constant. The coefficient of volume expansion of the gas is

(A) T1

(B) T2

(C) T3

(D) T4

Ans. (D)Sol. PT2 = C

using PV = nRT in PT3 = C

CTV

nRT 3

T4 V

Differentiating we get

4T

dT =

VdV

Coefficient of volume expansion () = dTdV

V1

= T4

21. Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen attemperature To, while box B contains one mole of helium at temperature (7/3)To. The boxes are then put intothermal contact with each other, and heat flows between them until the gases reach a common final temperature.(Ignore the heat capacity of boxes and table). Then, the final temperature of the gases, Tf in terms of T0 is :

(A) 0f T

73

T (B) 0f T

37

T (C) 0f T23

T (D) 0f T25

T

Ans. (C)Sol Equating internal energy,

1 oRT2

5 + 1 ffo RT

2

5RT

2

31T

3

7R

2

3

of T

2

3T


22. One mole of an ideal gas at a temperature T1 expands slowly according to the law 2V

p= constant. Its final

temperature is T2. The work done by the gas is

(A) 3R

(T1 T2) (B) 2R(T2 T1) (C) 2R

(T2 T1) (D) 3R

(T2 T1)

Ans. (D)

23. In the shown arrangement if f1, f2 and T be the frictional forces on 2 kg block, 3kg block and tension in thestring respectively, then their values are:

(A) 1 N, 6 N, 10 N (B) 2 N, 6 N, 5 N (C) 2 N, 6 N, 13 N (D) None of theseAns. (B)Sol. FBD

a = 2s/m1

52�7�14

14 � T � 6 = 3 (1)

T = 5

211

83kg62

2kg

So f1 = 2 N, f2 = 6 N, T = 5 N

24. Figure shows a plane mirror on which a light ray is incident. If the incident light ray is turned by 10º and the mirror

by 20º, as shown, then what will be the angle turned by the reflected ray.

(A) 10º clockwise (B) 20º clockwise (C) 30º clockwise (D) 50º clockwise

Ans. (C)Sol. Angle turned by the reflected ray = 2 (20º) � (10º) = 30º clockwise.

25. A rod of length 5 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a waythat the end farther from the pole is 20 cm away from it. Find the length of the image.(A) 5 cm (B) 10 cm (C) 20 cm (D) infinitely large

Ans. (B)

Sol.

So length of the image will be 10 cm.


26. A man uses a concave mirror for shaving. He keeps his face at a distance of 20 cm from the mirror andgets an image which is 1.5 times enlarged. What will be the radius of curvature mirror.(A) 15 cm (B) 30 cm (C) 60 cm (D) 120 cm

Ans. (D)

Sol. m = � uv

= 1.5 v = � 1.5 u

v = � 1.5 × (� 20) = 30 cm

20�

1301

f1

= 60

3�2f = � 60 cm.

R = 2f = 120 cm

27. A point object is kept in front of a plane mirror. The plane mirror is performing SHM of amplitude 1 cm. The planemirror moves along the x-axis and x- axis is normal to the mirror. The amplitude of the mirror is such that theobject is always infront of the mirror. The amplitude of SHM of the image is(A) zero (B) 2 cm (C) 4 cm (D) 1 cm

Ans. (B)

Hint :

(i) (ii)From figure (i) and (ii) it is clear that if the mirror moves distance �A� then the image moves a distance �2A�.

Sol. Therefore Amplitude of SHM of image = 2A

28. A plane mirror is moving with velocity k�3j�5i�4 A point object in front of the mirror moves with a velocity

k�5j�4i�3 . Here k� is along the normal to the plane mirror and facing towards the object. The velocity

of the image is :

(A) k� 5j�4i�3 (B) k�11j�4i�3 (C) k�11j�4i�3 (D) k� j�4i�3

Ans. (D)

Sol. k�5j�4i�3V0

2

VVV 0

m

k�3 = 2

Vk�5

VI = k� j�4i�3

29. A square ABCD of side 0.5 mm is kept at distance 15 cm infront of the concave mirror as shown in the figure. Thefocal length of the mirror is 10 cm. The length of the perimeter of its image will be(nearly):

(A) 8 mm (B) 2 mm (C) 12 mm (D) 6 mmAns. (D)


Sol. v = f�u

uf =

1015�

)10(�)15(�

= � 30 cm, m = �

uv

= � 2 AAB = CD = 2 × 0.5 = 1 mm

NowBC

CB =

ADDA

= 2

2

u

v = 4 BC = AAD = 2 mm

Perimeter length = 1 + 1 + 2 + 2 = 6 mm Ans.

30. Two blocks each of mass m lie on a smooth table. They are attached to two other masses as shown in thefigure. The pulleys and strings are light. An object O is kept at rest on the table. The sides AB and CD ofthe two blocks are made reflecting. The acceleration of two images (first image formed by both mirror)formed in those two reflecting surfaces w.r.t. each other is:

(A) Zero (B) 3g/2 (C) 3g (D) 17 g/6Ans. (C)

Sol. Acceleration of block AB = mm3

mg3

= 43

g ; acceleration of block CD = mm3

mg3

= 43

g

Acceleration of image in mirror AB= 2 acceleration of mirror

=

4

g3 .2 =

23

g

Acceleration of image in mirror CD =

4g3

.2 = 23

g

Acceleration of the two image w.r.t. each other =

2

g3��

2

g3 = 3g

SECTION - IITrue & False Type

This Section Contains 10 questions. Each question is either true (T) or false (F).

31. If a particle is moving under the action of force F = � kx3 , where k is a positive constant and x is displacementfrom mean position then particle will perform oscillatory motion along x-axis about origin.

Ans. (T)

32. For a particle performing SHM, its speed decreases as it goes away from the mean position.Ans. (T)

33. In a stationary wave, all the particles (other than node) of the medium vibrate either in phase or in oppositephase.

Ans. (T)

34. Coherent sources are the sources which are identical to each other and phase difference between the wavesfrom the two sources meeting at a point in the medium varies with time.

Ans. (F)


35. The number of nodes and antinodes are equal in a standing sound wave in a pipe, then the pipe has bothends open.

Ans. (F)Sol. If number of nodes = number of antinodes then pipe must be open at one end only. Hence false.

36. If change in volume (V) = 0 in a process then work done by the gas must be zero.Ans. (F)Sol. W.D. is path function even when V = 0 then W may not be zero.

37. Temperature of an ideal gas is increased from 70 ºC to 140 ºC. The average kinetic energy of each molecule will

doubled.Ans. (F)

Sol. K.E. = 2f

kT where T is in kelvin.

38. A convex mirror always forms virtual image for all positions of real object.Ans. (T)

39. If a parallel beam of light incident from rarer medium on a plane interface is refracted into denser medium,then the cross-section area of refracted beam cannot be less than that of incident beam.

Ans. (T)Sol. As the refracted beam bends towards normal its cross-section area will increase. Hence true.

40. When a concave mirror is held under water, its focal length increases.Ans. (F)Sol. The focal length of a concave mirror depends only on its radius of curvature.

PART-II (CHEMISTRY)

Atomic masses : [H = 1, D = 2, Li = 7, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27, Si = 28,P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40, Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65, As = 75,Br = 80, Ag = 108, I = 127, Ba = 137, Hg = 200, Pb = 207]


This section contains 30 multiple choice questions. Each question has choices (A), (B), (C) and (D), outof which ONLY ONE is correct.

41. Examine the T vs composition graph for two miscible liquids as shown, then mark the incorrect statement.

(A) Line '1' represents T vs composition of vapour phase.

(B) Line '2' represent T vs composition of liquid phase.

(C) Line '1' represents the vapour pressure of more volatile component only.

(D) A is the more volatile component.Ans. (C)


42. Which of the following is not a colligative property(A) Osmotic pressure (B) Elevation in B.P.(C) Vapour pressure (D) Depression in freezing point

Ans. (C)

43. Which of the following is true when components forming an ideal liquid solution are mixed.(A) H

mix. =V

mix. = 0 (B) H

mix. >V

mix.(C) H

mix. <V

mix.(D) H

mix. =V

mix. = 1

Ans. (A)

44. A water sample is collected in a container at 300 K and above it a mixture of gases A and B is present as shownin diagram . Find the total pressure on the surface of liquid if the water sample is found to contain 0.0001 and0.0002 mole fractions of gases A and B respectively .

(KH)A = 2 × 102 atm(KH)B| = 4× 102 atm aqueous tension of water at 300 K = 0.05 atm(A) 1.05 atm (B) 1 atm (C) 0.25 atm (D) 0.15 atm.

Ans. (D)Sol. Apply Henry's law

PA = 2 × 102 × 0.0001 = 0.02 atm

PB = 4× 102 × 0.0002 = 0.08

total pressure = 0.02 + 0.08 + 0.05 = 0.15 atm

45. The vapour pressure of a pure liquid A is 40 mmHg at 310 K. The vapour pressure of this liquid in a solution withliquid B is 32 mmHg. Mole fraction of A in the solution, if it obeys Raoult�s law is :

(A) 0.8 (B) 0.5 (C) 0.2 (D) 0.4Ans. (A)Sol. p

A = X

A p°

A

32 = XA 40

XA =

4032

= 0.8.

46. Two liquids X and Y are perfectly immiscible. If X and Y have molecular masses in ratio 1:2, the total vapourpressure of a mixture of X and Y prepared in weight ratio 2:3 should be (P

X0 = 400 torr, P

Y0 = 200 torr)

(A) 300 torr (B) 466.7 torr (C) 600 torr (D) 700 torrAns. (C)Sol. For immiscible solution = P

T = P

A0 + P

B0 = 400 + 200 = 600.

47. The Van't Hoff factor i for a 0.2 molal aqueous solution of urea is :(A) 0.2 (B) 0.1 (C) 1.2 (D) 1.0

Ans. (D)

48. A solution is prepared by mixing 0.01 mol of NaCl , 0.02 mole of glucose and 0.02 mol AlCl3 (33.33 % dissoci-

ated) in 360 g of water. If the solution contain no solid residue, then RLVP will be -(A) 1/250 (B) 1/125 (C) 1/251 (D) 1/502

Ans. (C)Sol. NaCl i = 2

glucose i = 1

AlCl3 i = 1+(4 � 1) ×

31

= 1 +3× 31

= 2

total moles solute particles = 2 × 0.01

+ 1× 0.02

+ 2× 0.02 = 0.02 + 0.02 + 0.04 = 0.08


total moles of solvent molecules = 18360

= 20

RLVP = Xsolute

= 08.020

08.0

= 08.20

08.0 =

20088

= 2511

49. PA and P

B are the vapour pressures of pure liquid components, A and B, respectively of an ideal binary solution.

If XA and Y

A represents the mole fraction of component A in liquid and vapour phase respectively, the total

pressure (P) of the solution will be given by :

(A) P = PA + X

A (P

B � P

A) (B) P =

B

B

A

A

PY

PY

(C) P =PB + X

A (P

B � P

A) (D)

P1

=B

B

A

A

PY

PY

Ans. (D)

50. 1 mole of an ideal gas is expanded isothermally and reversibly from 10L to 100L. Which of the following iscorrect for the process :(A) T 0 (B) H 0(C) heat supplied (q) 0 (D) E 0

Ans. (C)Sol. For an isothermal process, T = E = H = 0.

51. Heat of hydrogenation of ethene is x1 and that of benzene is x2. Hence, resonance energy of benzene is :(A) x1 � x2 (B) x1 + x2 (C) 3x1 � x2 (D) x1 � 3x2

Ans. (C)Sol. CH2 = CH2 + H2 CH3�CH3 H = X1

So, Hydrogenation energy of benzene sholud be 3x1H calculated = 3x1

So, Resonance energy = [ 3x1 � x2]

52. The species which by definition has zero standard molar enthalpy of formation at 298 K is :(A) CO

2 (g) (B) D

2(g) (C) O

3 (g) (D) P

4 (red)

Ans. (B)Sol. Standard molar enthalpy of formation (Hº

f) of element in their stable state of agregation is zero.

Hºf (D

2, g) = 0

53. Standard enthalpy of vapourisation vap

Hº for water at 100ºC is 40.66 kJ mol�1. The internal energy of vaporisation

of water at 100ºC (in kJmol�1) is :(A) + 37.56 (B) � 43.76 (C) + 43.76 (D) + 40.66

Ans. (A)Sol. H = E + n(g) RT

40.66 × 1000 = E + (1) × 8.314 × 373.

E = 37.56 kJ mol�1

54. According to molecular orbital theory , the paramagnetism of O2 molecule is due to the presence of :

(A) unpaired electrons in the bonding molecular orbital.(B) unpaired electrons in the antibonding molecular orbital.(C) unpaired electron in the bonding molecular orbitals.(D) unpaired electrons in the antibonding molecular orbitals.

Ans. (D)Sol. According to MOT the O

2 contains two electrons in 2p antibonding molecular orbitals as given below :

O2 (16) ; 1s2 *1s2 2s2 *2s2 2pz2 2px2 2py2 *2px1 *2py1


55. The pair of species with the same bond order is :(A) O

22� , B

2(B) O

2+ , NO+ (C) NO, CO (D) N

2, O

2

Ans. (A)Sol. Both O

22� and B

2 had bond order equal to 1.

56. The IUPAC name of the following compound is :

(A) 4-Bromo-3-cyanophenol (B) 2-Bromo-5-hydroxybenzonitrile(C) 2-Cyano-4-hydroxybromobenzene (D) 6-Bromo-3-hydroxybenzonitrile

Ans. (B)

Sol.

2-Bromo-5-hydroxybenzonitrile (�CN group gets higher priority over �OH and �Br)

57. In which of the following first resonating structure is more stable than the second ?

(A)

(B)

(C) CH2=CH�NH2

(D)

Ans. (C)

58. Which will be the least stable resonating structure :

(A) CH2 = CH � � � O � CH3 (B) 2HC

� � CH = CH � OCH3

(C) 2HC

� CH = CH � CH = � CH3 (D) CH2 = CH � � CH = � CH3

Ans. (A)Sol. Negative charge and lone pair on adjacent atom will increase potential energy and decrease stability.

59. The hybridisation of nitrogen in (pyrrole) is

(A) sp3 (B) sp2 (C) sp (D) Cann't be predictedAns. (B)


60. At how many carbons the negative charge of the following anion can reach through resonance (including thecarbon at which � ve charge is present in the given structure).

(A) 7 (B) 6 (C) 5 (D) 8Ans. (C)

Sol.

Only 5 carbons.

61. Resonance is not possible in :

(A) (B) (C) CH2 = CH � Cl (D)

Ans. (A)

62. + M and + I both effects are shown by :

(A) (B) (C) (D) � C (CH3)

3

Ans. (C)

63. Which group shows + M effect when attached with a benzene ring :

(A) (B) � COOH (C) � Ph (D) � CF3

Ans. (C)

64. Least contributing resonating structure of nitroethene is :

(A) (B) (C) (D)

Ans. (C)Sol. In (C) two positive charge on adjacent position.

65. Which of the following statements is incorrect ?(A) The energy of resonance hybrid is always less than that of any resonating structure.(B) The resonance energy is the difference between the enthalpies of formation of the molecule and the resonating

structure having maximum energy.(C) The resonance structures are hypothetical structure and they do not represent any real molecule.(D) In delocalized structure of benzene the -charge cloud is spread equally above and below the plane of

molecule.Ans. (B)Sol. Self explanatory.

66. Which of the following acid has the lowest value of acid dissociation constant Ka:

(A) CH3CHFCOOH (B) FCH

2CH

2COOH (C) BrCH

2CH

2COOH (D) CH

3CHBrCOOH

Ans. (C)Sol. Acidic strength Dissociation constant of acid. In given acids order of acidic strength

CH3 CHFCOOH > CH

3CHBrCOOH > FCH

2CH

2COOH > BrCH

2CH

2COOH

Hence in these smallest dissociation constant for BrCH2 �CH

2COOH


67. Which of the following is not acceptable as resonating structure :

(A) (B) (C) (D) None of these

Ans. (B)

68. The strongest +M group is :

(A) �OH (B) �OCH3 (C) (D) �ONa

Ans. (D)

69. Which of the following is the major contributor to the resonance hybrid of CH3COOCH

3 ?

(A) CH � C � O � CH3 3

-

�

+

O

(B) CH � C = O � CH3 3

-�

+

O

(C) CH � C � O � CH3 3

-

�

+O

(D) CH � C = O � CH3 3

-

�

+O

Ans. (B)Sol. Due to complete octet and maximum number of covalent bond.

70. Which of the following compound have delocalisation of -electrons ?

(A) (B)

(C) CH3CH

2NH CH

2 CH = CH

2(D) CH

2 = CH � CH

2 � CH = CH

2

Ans. (B)

Sol.


This Section Contains 10 questions. Each question is either true (T) or false (F).

71. Acetic acid undergoes association in benzene. The molar mass of acetic acid, determined by elevation ofboiling point measured in benzene solvent is always higher than its normal molar mass.

Ans. [T]

Sol. Mobserved = i

M ltheoretica ; i < 1

72. In case of minimum boiling azeotrope, the boiling point of the azeotropic mixture is less than the boiling point ofany of the two pure component.

Ans. [T]

73. Among the species K2O

2 , KO

2 , O

2 , O

2+ [AsF

6] � ,compound K

2O

2 will have the largest O � O bond length.

Ans. [T]

Sol. O2� B.O = 1

21

, O2 B.O = 2 ,O

2+ = 2

21

, O2

2 � = 1

More is the bond order less is the bond length.


74. Less work is done on the gas in adiabatic reversible compression than in isothermal reversible compression, ifcompression is performed between same initial and final volumes and same initial pressure.

Ans. [F]

75. In the compound NO[BF4], nitrogen oxygen bond length is lower than nitric oxide (NO) and B�F bond length is

higher than in BF3.

Ans. [T]

76. In propenal (CH2 = CH � CH = O) partial displacement of electrons and complete displacement of

electrons take place in same direction.Ans. [T]

77. � N = O group shows � M effect only.

Ans. [F]

Sol. group can show both � M as well as + M effect.

78. Conjugation is not present in CH2 = C = O.

Ans. [F]Sol. Lone pair at O and C = C are in conjugation.

79. Although inductive effect is observed in (hydroquinol) but it is a non-polar compound.

Ans. [F]Sol. It is a polar compound.

80. The most stable resonating structure of is .

Ans. [F]

Sol. this is least stable structure due to incomplete octet and charge species.

PART-III (MATHEMATICS)


This section contains 30 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONLY ONE is correct.

81. The complete set of values of for which equation ||x|2 � 4 |x| + 3| + 3 = has no solution, is(A) (�, 0) (B) (�, 3) (C) (0, 3) (D) (3, 4)

Ans. (B)Sol. ||x|2 � 4 |x| + 3| = � 3

for no solution � 3 < 0

< 3

82. Domain of f(x) = cos�1 x + cot�1 x + cosec�1 x is(A) [� 1, 1] (B) R (C) (� ] [1, ) (D) {� 1, 1}

Ans. (D)Sol. �1 x 1 ...(1)

x R ...(2)x �1 or x 1 ...(3)

By (1) (2) (3) x {�1, 1}


83. If cos )))3

1(cotsin(tan( 11 = y, then

(A) y = 5

4(B) y =

5

2(C) y =

7

2(D) y2 =

10

11

Ans. (C)

Sol. cos )))3

1(cotsin(tan( 11 = y

= cos (tan�1 (sin 3

)) = cos

23

tan 1� = 7

2

84. If sgn(sin�1(x � 1)) = �1, then (where sgn(.) represents signum function)(A) x (�, 1) (B) x [0, 1] (C) x [0, 1) (D) x [0, 2]

Ans. (C)Sol. �1 x � 1 1 and sin�1 (x � 1) < 0

0 x 2 and x � 1 < 0

x [0, 1)

85. The complete solution set of the inequality [cos�1x]2 � 6 [cos�1 x] + 9 0, where [.] denotes greatestinteger function, is(A) (� , cos 3] (B) [cos 3, cos 2] (C) [cos 3, cos1] (D) none of these

Ans. (D)

Sol. ([cos�1x] � 3)2 0 3]x[cos 1

4xcos3 1 4xcos3 1

3cosx1

86. If |x2 � 2x � 8| + |x2 + x � 2| = 3 | x + 2|, then the set of all real values of x is

(A) [1, 4] {�2} (B) [1, 4] (C) [�2, 1] [4,) (D) (�, �2] [1, 4]Ans. (A)Sol. |x2 � 2x � 8| + |x2 + x � 2| = 3 | x + 2| and (x2 + x - 2) - (x2 - 2x - 8) = 3x + 6 = 3(x+2)

(x2 - 2x - 8) (x2 + x - 2) 0i.e. (x - 4) (x + 2) (x + 2) (x - 1) 0

Solution set is [1,4] {-2}

87. Values of �x� satisfying the inequation cos 2x 23

are

(A)

12,0 (B)

6,0 (C)

2

3, (D)

2,

12

11

Ans. (A)

Sol. cos 2x 23

2x

6,0

2,

6

11

x

12,0

,

1211

1213

,

2,

1223


88. The number of solutions of the equation 2{x}2 � 5 {x} + 2 = 0 are (where {.} denotes the fractional partfunction)(A) 0 (B) 1 (C) 2 (D) infinite

Ans. (D)Sol. 2{x}2 � 5{x} + 2 = 0

So {x} = 21

,2

but {x} 2

So {x} = 21

x has infinite values.

89. The value of [e2] � [� ]2 is, where [.] denotes greatest integer function(A) 17 (B) 6 (C) �10 (D) �9

Ans. (D)Sol. [e2] � [� ]2

here e2 7.29and 3.14 [7.29] � [� 3.14]2

= 7 � (�4)2 = �9

90. The roots of quadratic equation (b � c) x2 + (c � a) x + (a � b) = 0 are

(A) cbac

, 1 (B)

cbba

, 1 (C)

bacb

, 1 (D)

baac

, 1

Ans. (B)Sol. x = 1 is root

Let other root =

Product of the roots = (1) () = cbba

roots are 1,

cbba

91. Which of the following graph represents y = a x2 + b

x + c, (a 0) when a > 0, b < 0 & c < 0 ?

(A) (B) (C) (D)

Ans. (B)Sol. a > 0 & c < 0 is satisfied by (B) only [ f(0) = 0 & a > 0]

Further in (B)

�a2

b > 0 b < 0 [ a > 0].

92. The complete set of values of a for which the quadratic equation x2 + 2ax + a2 � 1 = 0 has atleast one root

negative, is given by(A) (1, ) (B) (�1, ) (C) (�, 1) (D) (�, �1)

Ans. (B)Sol. (x + a)2 = 1 x = �a � 1, �a + 1

since atleast one root is negative �a � 1 < 0 a > �1


93. Let and be the roots of the quadratic equation x2 � 1154 x + 1 = 0, then the value of 44 is equal to

(A) 4 (B) 5 (C) 6 (D) 8Ans. (C)Sol. + = 1154 and = 1

2 = + + 2 = 1154 + 2 = 1156 = (34)2 = 34

Again (1/4 + 1/4)2 = + 2()1/4 = 34 + 2 = 36

1/4 + 1/4 = 6

94. The value of sin( +) sin ( � ) cosec2 is equal to (where n,n )(A) � 1 (B) 0 (C) sin (D) none of these

Ans. (A)Sol. (� sin) (sin) × cosec2 = � 1

95. If A + B = 225°(A (2n + 1)180º, B (2n + 1)180º, n ), then the value of

Acot1Acot

.

Bcot1Bcot

is

(A) 2 (B) 21

(C) 3 (D) 31

Ans. (B)Sol. cot (A + B) = cot 225° = 1

BcotAcot

1�BcotAcot

= 1

cot A cot B = 1 + cot A + cot B

Now BcotAcotBcotAcot1

Bcot.Acot

= )BcotAcot1(2

BcotAcot1

=

21

96. Domain of the function f(x) = sin�1 sin

1x

xn is

(A) x (�, �1)

,

1e

1(B) x (�, �1) ,0

(C) x (�, e1

e

) (D) none of these

Ans. (B)

Sol.1x

x

> 0

x (�, �1) ,0

97. If sin = 1, then the general value of is

(A) = (4n + 1)2

, n (B) = (4n � 1)2

, n

(C) = (2n + 1)2

, n (D) none of these

Ans. (A)


Sol. sin = 1

= n + (�1)n 2

Case-(i) n is evenn = 2m

= 2m + 2

= (4m + 1)2

Case -(ii) n is odd n = 2m +1

= (2m + 1) � 2

= (4m + 1)2

98. The roots of the equation 2x5x21

52�1

2041

= 0 are

(A) �1, �2 (B) �1, 2 (C) 1, �2 (D) 1, 2Ans. (B)

Sol.2x5x21

52�1

2041

= 0

�30x2 + 30x + 60 = 0 x2 � x � 2 = 0

x = 2, �1

99. If7x4x1x23

x32xx7x2

4x2x73x2x

2

2

2

= ax6 + bx5 + cx4 + dx3 + ex2 + fx + g, then the value of g is

(A) 2 (B) 1 (C) 2 (D) none of theseAns. (D)

Sol.7x4x1x23

x32xx7x2

4x2x73x2x

2

2

2

= ax6 + bx5 + cx4 + dx3 + ex2 + fx + g

put x = 0

71�3

027

423

= g

3(14) � 2 (49) + 4 (�7 �6) = g

42 � 98 � 52 � 108 = g

100. The system of equations x + y + z = 2, 3x � y + 2z = 6, 3x + y + z = �18, has

(A) a unique solution (B) no solution(C) an infinite number of solutions (D) zero solution as the only solution

Ans. (A)Sol. x + y + z = 2

3x � y + 2z = 6

3x + y + z = �18

D = 113

21�3

111

= 1(�1�2) �1(3 � 6) + 1(3 + 3)

= �3 + 3 + 6 = 6

so it has unique solution


101. The complete solution set of the inequation 3|x|1

< 41

is

(A) (B) (1, 1) (C) (�, �1) (1, ) (D) none of theseAns. (C)

Sol. 3|x|1

< 41

|x| + 3 > 4 |x| > 1 x (�, �1) (1, )

102. Graph of y = {x} + {� x} in the interval [�1, 2] is (where {.} denotes fractional part function)

(A) (B)

(C) (D)

Ans. (D)

Sol. y = {x} + {�x} =

x

x

1

0

103. Graph of y = 1�x6x9� 2 is same as the graph of

(A) y = �9x2 + 6x � 1 (B) y = 3x � 1 (C) y = |3x � 1| (D) |y| + |3x � 1| = 0

Ans. (D)

Sol. y = 2)1�x3(�

Domain is �(3x � 1)2 0 (3x � 1)2 0 x = 31

y = 0

104. Solution set of the inequation 9x6�x

4x4�x2

2

0 is

(A) x (�, ) (B) x R � {2, 3} (C) x R � {2} (D) none of theseAns. (D)Sol. x2 � 6x + 9 > 0

(x � 3)2 > 0 x R � {3}

Solution is x R � {3}

105. If sin + cosec = 2, then value of (sin)2013 + (cosec)2013 is(A) 2 (B) �2 (C) 2011 (D) 2013

Ans. (A)Sol. sin + cosec = 2 sin = 1

(sin)2013 + (cosec)2013 = 2


106. Number of integral values of x in the interval [1, 5] satisfying the equation }x{]x[sin= 0 (where [.] and {.} denotes

greatest integer function and fractional part function respectively) are(A) 0 (B) 3 (C) 5 (D) infinite

Ans. (A)

Sol. }x{]x[sin= 0

It is true x R � I {x} = 0 at x I

107. Number of integers satisfying the equation [x] + [2x] + [3x] = {x} + {2x} + 2x2 (where [.] and {.} denotes greatestinteger function and fractional part function respectively) are(A) 0 (B) 1 (C) 2 (D) 3

Ans. (C)Sol. Since x is integer, {x} and {2x} = 0

x + 2x + 3x = 2x2

2x2 � 6x = 0 x = 0, 3

108. The domain of the function f(x) = sin�1(x +[x]), where [.] denotes the greatest integer function, is(A) [0, 1) (B) [�1, 1] (C) (�1, 0) (D) none of these

Ans. (A)Sol. �1 x + [x] 1, if �1 x < 0 then x + [x] < � 1

and if x 1 then x + [x] > 1 0 x < 1

109. If [x1 + x

2 + x

3 + x

4] = p and [x

1] + [x

2] + [x

3] + [x

4] = q, where x

1, x

2, x

3, x

4 R and [.] denotes the greatest

integer function, then the maximum value of (p � q) is

(A) 0 (B) 1 (C) 2 (D) 3Ans. (D)Sol. Let x

1 = I

1 + f

1

x2 = I

2 + f

2

x3 = I

3 + f

3

x4 = I

4 + f

4

p = [I1 + I

2 + I

3 + I

4 + f

1 + f

2 + f

3 + f

4]

p = I1 + I

2 + I

3 + I

4 + [ f

1 + f

2 + f

3 + f

4]

p = q + [f1 + f

2 + f

3 + f

4]

p � q = [f1 + f

2 + f

3 + f

4]

0 f1 < 1

0 f2 < 1

0 f3 < 1

0 f4 < 1

0 f4 < 1

0 f1 + f

2 + f

3 + f

4 < 4

Maximum value of [f1 + f

2 + f

3 + f

4] = 3

110. The number of solutions of equation |4 � |x|| = 4�|x| are(A) 1 (B) 2 (C) 3 (D) 4

Ans. (D)

Sol.



This Section Contains 10 questions. Each statement is either true (T) or false (F).

111. loga (x . y) = log

ax + log

ay is true for all x, y R.

Ans. (F)Sol. log

a (x · y) = log

ax + log

ay is true for x, y R+ statement is false.

112. If ,,() are roots of 6x2 + 11x + 3 = 0 then cos� 1exist but cos� 1does not exist.Ans. (T)

Sol. = �31

, = � 23

< � 1

Hence the result

113. If we consider only the principal values of the inverse trigonometric functions then the value of

tan

17

4sin

25

1cos 11

is 292

.

Ans. (F)Sol. Let y = tan ( � )

= cos�1

25

1 ; = sin�1

17

4

So y = 4.71

47

=

293

.

114. If and are roots of equation x2 + 2x + 4 = 0, then 33

11

=

41

Ans. (T)Sol. + = �2 = 4

3

3

33

33

33 )(

)(3)(11

= 41

6416

64248

115. If b2 4ac, then roots of quadratic equation ax2 + bx + c = 0 are real.Ans. (F)

116. The number of integral values of , for which equation 4cos x + 3 sin x = 2 + 1 has solutions are 6.Ans. (T)Sol. � 5 2 + 1 5

� 6 2 4� 3 2 number of solution is 6.

117. If cos = k, 0 < k < 1 and does not lie in the first quadrant, then the value of the cosecis 2k1

1

.

Ans. (F)Sol. cos = k, 0 < k < 1

0 < cos < 1 & Ist quadrant IVth quadrantand in IVth quadrant cosec is -ve

cosec = � 2k�1

1


118. If graph of y = f(x) is ,

then the graph of y = )x(f1

is

Ans. (F)

Sol. Correct graph is

119. If �1 {x} 0 then y = cosec(x) is not defined. (where {.} denotes fractional part function)Ans. (T)Sol. �1 {x} 0

{x} = 0 x . y = cosec(x) not defined.

120. The domain and range of arcsecx is R � (�1, 1) and [0, ] � {/2} respectively.Ans. (T)Sol. True by definition

PAPER-2

PART-I (PHYSICS)


This section contains 5 multiple choice questions. Each question has choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.

1. A cylindrical piston of mass M slides smoothly inside a long cylinder closed at one end as shown in figureenclosing a certain mass of a gas. The cylinder is kept with its axis horizontal. If the piston is disturbed slightlyfrom its equilibrium position, it oscillates simple harmonically. The period of oscillation will be (process isisothermal)

(A) PAMh

2T (B) PhMA

2T (C) PAhM

2T (D) MPhA2T

Ans. (A)


Sol.

force = (P � P0) A = Ma

by P1V

1 = P

2V

2

P0hA = P(h � x) A P =

x�hhP0

So force =

0

0 P�x�h

hP

MA

= a

a = )x�h(MAxP0

Mh

AxP0 ; (for small x)

T = 2 PAMh

2. An object of specific gravity ( = 2) is hung from a thin steel wire. The fundamental frequency for transversestanding waves in the wire is 300 Hz. The object is immersed in water so that one half of its volume issubmerged. The new fundamental frequency in Hz is : (

w = 1000 kg/m3)

(A) 3300 (B) 3150 (C) 150 (D) 3

300

Ans. (B)

Sol. If T = mg = Vg

f = 2

1

T = 300 .....(1)

Now T´ = mg � fb = Vg �

2V

g.

T´ = Vg

2

1�2

f´ = 2

1 2

)1�2(Vg

.......(2)

f´f

= ½

21�2

f´ = 300

½

2

1�2

= 3150

43

3002/1

.

3. In a Hall, a person receives direct sound waves from a source 120m away. He also receives wave from the samesource which reach him after being reflected from the 25m high ceiling at a point half way between them. The twowaves interfere constructively for wave length (in meters).(A) 10, 10/2, 10/3, 10/4 (B) 20, 20/3, 20/5, 20/7,...........(C) 30, 20, 10,............ (D) 10, 10/3, 10/5,10/7...........

Ans. (A)


Sol. Path difference (S) between direct and reflected wave = 130 � 120 = 10 m

for so constructive interference S = n

= n

10nS

(n = 1, 2, 3, .......)

= 10, 4

10,

310

,2

10 .....

4. An object is placed 30 cm (from the reflecting surface) in front of a block of glass 10 cm thick having itsfarther side silvered. The final image is formed at 25 cm behind the silvered face. The refractive index ofglass is :

(A) 23

(B) 34

(C) 132200

(D) 45

Ans. (B)

Sol.

Distance of 1 from refracting surface = 20 Distance of 2 from reflecting surface = 20µ + 10

Distance of 2 from refracting surface = 20 + 20 Distance of 3 from refracting surface

=

2020= 10 + 25

=

20 + 20 = 35

= 1520

= 34

5. A wave travels on a light string.The equation of the wave is Y = A sin (kx � t + 30°).It is reflected from a heavy

string tied to an end of the light string at x = 0.If 64% of the incident energy is reflected the equation of thereflected wave is(A) Y = 0.8 A sin (kx � t + 30° + 180°) (B) Y = 0.8 A sin (kx + t + 30° + 180°)

(C) Y = 0.8 A sin (kx + t � 30°) (D) Y = 0.8 A sin (kx + t + 30°)

Ans. (C)Sol. As y = A sin (Kx � t + 30º) for incident wave

Now for reflected wave : Energy Amp2

Y = 0.8 A sin (�Kx � t + 30 + 180) Y = 0.8 A sin (�Kx � t + 210) Y = �0.8 A sin (kx + t �210)

Y = �0.8 A sin [Kx + t � 30 � 180]

Y = 0.8 A sin [180 � (Kx + t � 30)]

Y = 0.8 A sin (Kx + t � 30)


SECTION - IIMultiple Correct Answers Type

This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A),(B), (C) and (D), out of which ONE OR MORE is/are correct.

6. A cylindrical tube, open at one end and closed at the other, is in acoustic unison (resonance) with anexternal source of sound of single frequency held at the open end of the tube, in its fundamental note. Then:(A) the displacement wave from the source gets reflected with a phase change of at the closed end(B) the pressure wave from the source get reflected without a phase change at the closed end(C) the wave reflected from the closed end again gets reflected at the open end(D) the wave reflected from the closed end does not suffer reflection at the open end

Ans. (A), (B), (C)Sol. When sound wave is reflected from rigid end displacement wave get extra phase at and pressure wave get no

extra phaseSo option A, B, C are correct .

7. When the temperature of a copper coin is raised by 80 oC, its diameter increases by 0.2%,(A) percentage rise in the area of a face is 0.4%(B) percentage rise in the thickness is 0.4%(C) percentage rise in the volume is 0.6%(D) coefficient of linear expansion of copper is 0.25x10-4 / oC.

Ans. (A), (C), (D)

Sol.AA

× 100 =

A2

× 100

% increase in Area = 2 × 0.2 = 0.4

VV

× 100 = 3 × 0.2 = 0.6 %

Since l = l T

× 100 = T × 100 = 0.2

= 0.25 × 10�4 / ºC

8. One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its pressure

at A is P0. Choose the correct option(s) from the following :

(A) Internal energies at A and B are the same (B) Work done by the gas in process AB is P0V0 n 4

(C) Pressure at C is 4

P0 (D) Temperature at C is 4

T0

Ans. (A), (B)


Sol.

U = 2f

nRT,, where f,n,R are constants. Also temperature T is same at A & B.

UA = UB

Also, WAB = nRT0

i

f

V

V n = nRT0

0

0

V

V4 n = nRT0 n4 = P0V0 n4

So, answers are (A) & (B).

9. is the image of a point object O formed by spherical mirror, then which of the following statements is correct:(A) If O and are on same side of the principal axis, then they have to be on opposite sides of the mirror.(B) If O and are on opposite side of the principal axis, then they have to be on same side of the mirror.(C) If O and are on opposite side of the principal axis, then they can be on opposite side of the mirror as well.(D) If O is on principal axis then has to lie on principal axis only.

Ans. (A), (B), (D)

Sol.O

I = �

uv

If O and I are on same sides of PA . O

I will be positive which implies v and u will be of opposite signs.

Similarly if O and I are on opp. sides, O

I will be -ve which implies v and u will have same sign.

If O is on PA, I =

u

V (O) = 0 I will also be on. P.A.

SECTION - IIIComprehension Type

This section contains 3 paragraphs. Based upon each paragraph, 2 multiple choice questions have to beanswered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Paragraph for Question Nos. 10 to 11

Two particles A and B are performing SHM along x and y-axis respectively with equal amplitude and frequencyof 2 cm and 1 Hz respectively. Equilibrium positions of the particles A and B are at the co-ordinates (3 cm, 0cm) and (0 cm, 4 cm) respectively. At t = 0, B is at its equilibrium position and moving towards the origin, whileA is nearest to the origin and moving away from the origin.

10. During their motion the maximum distance between A and B is :(A) 3 cm (B) 4 cm (C) 5 cm (D) 7 cm

Ans. (D)

11. During their motion the minimum distances between A and B is :(A) 3 cm (B) 4 cm (C) 5 cm (D) 7 cm

Ans. (A)


Sol.(10 to 11)

At t = 0 Particle 2 is at point B and moving towards origin so displacementY = 4 � A sin tY = 4 � 2 sin tand displacement of particle 1 is X = 3 � A cos t

So distance between them = 22 YX

d2 = 29 � (16 sin t + 12 cos t) = 29 � 4 (4 sint + 3 cost) 29 � 20 sin(t + 37º)

So 2maxd = 49 d

max = 7

2mind = 9 d

min = 3


A standing wave is produced in a steel wire of mass 100 gm tied to two fixed supports at its ends. Thelength of the string is 2 m & strain in it is 0.4 %. The string vibrates in four loops. Assuming one end of thestring to be at x = 0, all particles to be at rest at t = 0 and maximum amplitude to be 3 mm :[Given : density of steel = 4 x 103 kg/m3, young�s modulus of steel =1.6 x 1011N/m2 ,2 = 10 ]

12. What will be the velocity of the travelling waves whose superposition is the given standing wave :(A) 200 m/sec (B) 100 m/sec (C) 400 m/sec (D) 800 m/sec

Ans. (C)

13. What will be wavelength & frequency of the travelling wave :(A) = 1 m , f = 400 Hz (B) = 1/2 m , f = 200 Hz(C) = 2 m , f = 400 Hz (D) = 4 m , f = 200 Hz

Ans. (A)Sol.(12 to 13)

Strain = LL

Stress = Y strain

= 1.6 × 1011 × 100

4.0= 6.4 × 108

V =

T =

ASA

V = 3

8

104

104.6

= 400 m/sec

2 = 2 = 1 m f = 400 Hz = 2f = 800

K = v

= 2



A concave mirror of radius of curvature 20 cm is shown in the figure. A circular thin disc of diameter 1 cm isplaced on the principle axis of mirror with its plane perpendicular to the principal axis at a distance 15 cm fromthe pole of the mirror. The radius of disc starts increasing according to the law r = (0.5 + 0.1 t) cm where t is timeis second.

14. The image formed by the mirror will be in the shape of a :

(A) circular disc (B) elliptical disc with major axis horizontal(C) elliptical disc with major axis vertical (D) distorted disc

Ans. (A)Sol. All dimensions of the disc are perpendicular to the principal axis. Hence all dimensions are equally magnified,

resulting in an image in the shape of a circular disc.

15. In the above question, the area of image of the disc at t = 1 second is :(A) 1.2 cm2 (B) 1.44 cm2 (C) 1.52 cm2 (D) none of these

Ans. (B)Sol.(14 to 15)

At t = 1 sec.r = 0.5 t + 0.1 t = 0.6 cm

m = uf

f

= 1510

10

= � 2

Radius of image = 2r = 1.2 cm Area of image = (1.2)2 = 1.44 cm2 .

SECTION - IVInteger Answer Type

This section contains 6 questions. The answer to each of the questions is asingle digit integer, ranging from 0 to 9. The appropriate bubble below therespective question number in the ORS have to be darkened.

16. One mole of a gas expands with temperature T such that its volume, V = kT2, where k is a constant. Thetemperature of the gas changes by 60º C then the work done by the gas is 250 y (in joule) then write the value

of y ? (R = 25/3 J/mol-K).Ans. 4

Sol. W = PdV

V = kT2 (given)

PdV = V

nRT 2kT dT

= 2kT

dTkT2nRT

PdV = 2nR dT

W = dTnR2PdV

W = 2nR T = 2 × 1 × R × 60

W = 120 R = 1000 J = 250 yy = 4


17. A uniform rope of length 20 m and mass 8 kg hangs vertically from a rigid support. A block of mass 2 kg isattached to the free end of the rope. A transverse pulse of wavelength 0.06 m is produced at the lower end of the

rope. If the wavelength of the pulse when it reaches the top of the rope is 510

y meter then write the value of y..

Ans. 3

Sol. At the bottom end

V =

mg = f ......(1)

Now at the top

V1 =

gmg

= f´ ......(2)

= M = mass of stringFrom equation (1) & (2)

´ = m

Mm =

282

= 5 = 510

3 m =

510

y m

y = 3

18. A particle executes simple harmonic motion (mean position is at origin) with an amplitude of 10 cm and timeperiod 6 s. At t = 0 it is at position x = 5 cm going towards positive x-direction. If the magnitude of the

acceleration of the particle at t = 4 sec is X

102

cm/sec2 then write the value of X .

Ans. 9Sol. A = 10 cm, T = 6 sec

So x = A sin

t

T2

= (10 cm) sin

t

62

= 10 cm sin

t

3

at t = 0, x = 5 cm 5 cm = 10 cm sin (0 + )

21

= sin ()

= 6

So, x = (10 cm) sin

6t

3

acceleration a = � 2 (10 cm) sin

6t

3

= �

2

3

× (10 cm) sin

64

3

= 9

102 =

X102

cm/s2

X = 9


19. The co-efficient of thermal expansion of a rod is temperature dependent and is given by the formula

= a T, where a = 21n2 °C�2 and T in ºC. If the length of the rod is at temperature 0 ºC, then find the

temperature (in °C) at which the length will be 2.Ans. 2

Sol. As ; d = dT

2d

= T

0

dTTa

n2 = a 2

T2

T =

2/14n

a

a4n

= 2

20. In the given figure rays incident on an interface would converge 2 cm below the interface if they continued tomove in straight lines without bending. But due to refraction, the rays will bend and meet some where else. Findthe distance of meeting point of refracted rays (in cm) below the interface, assuming the rays to be making smallangles with the normal to the interface.

Ans. 5

Sol. Apparent depth = 2

1

nn

× d =25

× 2 = 5 cm

21. A small object is placed at the centre of the bottom of a cylindrical vessel of radius 3 cm and height

3 3 cm filled completely with a liquid. Consider the ray leaving the vessel through a corner. Suppose thisray and the ray along the axis of the vessel are used to trace the image. If the apparent depth of the image

is y 3 cm then what will be the value of y. Refractive index of liquid = 3 .

Ans. 1

Sol.

Applying snells law : 3 sin 30º = 1 × sin r

r = 60º

tan r = A

AP 3 =

A3

AI = 3 cm = y 3 cm

y = 1


PART-II (CHEMISTRY)

Atomic masses : [H = 1, D = 2, Li = 7, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27, Si = 28,P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40, Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65, As = 75,Br = 80, Ag = 108, I = 127, Ba = 137, Hg = 200, Pb = 207]


This section contains 5 multiple choice questions. Each question has choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.

22. If SR = 100 torr and PR = 350 torr then the mole fraction of A in vapourphase and mole fraction of A in liquid phase respectively are :

(A) 103

, 92

(B) 97

, 103

(C) 92

, 107

(D) 92

, 103

Ans. (D)Sol. PQ = SR = P

A and PR = P

B

PT = P

A + P

B = 100 + 350 = 450 and P

B = Pº

BX

B

X'B (vapour phase) =

T

BPP

= 450350

= 97

350 = 500 XB

X'A = 1 � X'

B =

92

XA = 1 �

107

= 103

23. The enthalpy of combustion of H2, cyclohexene (C

6H

10) and cyclohexane (C

6H

12) are � 241, � 3800 and

� 3920 kJ per mol respectively. Heat of hydrogenation of cyclohexene is :

(A) � 121 kJ per mol (B) + 121 kJ per mol (C) + 242 kJ per mol (D) � 242 kJ per mol

Ans. (A)

Sol. + H2

H = [H of combustion of cyclohexane � (H of combustion of cyclohexene + H of combustionof H2)]

= �[� 3920 �( � 3800 � 241) ] kJ

= �[� 3920 + 4041] kJ

= �[ 121] kJ

= � 121 kJ

24. Which of the following molecular orbital has nodal planes perpendicular to each other ?(A) 2s (B) 2p

x(C) *2p

y(D) *2p

z

Ans. (C)Sol. * 2p

y has two nodal planes


25. The correct stability order of the following resonating structures is :

(A) (I) > (II) > (IV) > (III) (B) (I) > (III) > (II) > (IV)(C) (II) > (I) > (III) > (IV) (D) (III) > (I) > (IV) > (II)

Ans. (B)

Sol.

octet complete octet incomplete octet complete octet incomplete�ve charge on nitrogen �ve charge on nitrogen �ve charge on carbon �ve charge on carbon

26. The increasing order of + M effect of the following :

(A) I < II < III (B) I < III < II (C) II < III < I (D) III < II < I

Ans. (A)

SECTION - IIMultiple Correct Answers Type

This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A),(B), (C) and (D), out of which ONE OR MORE is/are correct.

27. Which of the following is/are correct order of difference of boiling point and freezing point of the aqueous solution.(A) 0.1 m K

4[Fe(CN)

6] > 0.1 m BaCl

2 > 0.1 m urea.

(B) 0.1 m Na3[Ag(S

2O

3)

2] > 0.2 m glucose > 0.1 m CH

3COOH (a weak acid).

(C) 0.1 m urea > 0.1 m NaCl > 0.1 m Na3PO

4

(D) 0.2 m glucose > 0.2 m CaCl2 > 0.1 m K

4[Fe(CN)

6].

Ans. (AB)Sol. T

B + T

� = ik

bm + ik

fm

TB � T

� � 100 = i(k

b + k

f) m

TB � T

� im


28. Consider an ideal solution of CHCl3 and CCl

4() . Given that vapour pressure of CHCl

3 and CCl

4 are 400 torr and

100 torr respectively . Which of the following is (are) true for an equimolar mixture of CHCl3 and CCl

4 as external

pressure is reduced over this mixture starting from one atmosphere?(A) It is completely in liquid state at one atmosphere pressure.(B) As pressure is reduced below 400 torr a bubble of pure CHCl

3 appears.

(C) As pressure is dropped to 250 torr both CHCl3 and CCl

4 are present in the bubble of vapour.

(D) At 150 torr no liquid mixture remains . it is completely vaporized.Ans. (ACD)

Sol.Z1

P1

100

1

400

1 =

2001

(1.25)

P = 160 torr

29. In which of the option/s the group/s is/are overall electron releasing when attached to a benzene ring.

(A) (B) �NO2 (C) (D)

Ans. (ACD)

30. Which of the following is/are resonating structures of diazomethane (CH2N

2).

(A) (B) (C) (D) all of these

Ans. (BC)Sol. Self explanatory.


This section contains 3 paragraphs. Based upon each paragraph, 2 multiple choice questions have to beanswered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Paragraph for Question Nos. 31 to 32Using the data given below :The standard enthalpy of formation of C

2H

2 (g) = 54 kJ/mole,

The internal energy change of the combustion of C6H

6(g) and C

2H

2(g) at 300 K are � 800 Kcal/mole and

� 300 Kcal/mole respectively. Assume that C6H

6 (g), C

2H

2 (g) and C

2H

6 (g) behave ideally.

Bonds Bond dissociation energies (kJ/mole)C�H 414C�C 347C=C 615H�H 435CC 962

(Take R = 2 cal K�1 mol�` ; 10 = 3.16; 1 cal = 4.2 J)

31. What is the enthalpy of polymerisation of C2H

2 (g) to C

6H

6 (g) per mole of C

6H

6 (g) produced ?

(A) � 100 K Cal/mol (B) � 101.2 K Cal/mol (C) �98.8 K Cal/mol (D) � 103.6 K Cal/mol

Ans. (B)


Sol. C6H

6 (g) +

215

O2 (g) 6CO

2 (g) + 3H

2O()

ru

1 = � 800 kcal/mole

C2H

2 (g) +

25

O2 (g) 2CO

2 (g) + H

2O()

ru

2 = � 300 kcal/mole

3C2H

2 (g) C

6H

6 (g)

ru = � 100 kcal/mole

rH =

ru + (�2) × 2 × 300 × 10�3

rH = � 100 � 1.2 = 101.2 kcal/mol.

32. What is the standard enthalpy of formation of C2H

6 (g) is :

(A) �171 kJ/mole (B) � 117 kJ/mole (C) �225 kJ/mole (D) �17 kJ/mole

Ans. (B)

Sol. C2H

2 (g) + 2H � H (g)

HH||

H�C�C�H||HH

(g) rHº

rHº =

CC + 2

C�H + 2

H�H � 6

C�H �

C�C

= � 171 kJ/mol

or rHº =

�Hº (C

2H

6, g) �

�Hº (C

2H

2, g)

or �Hº (C

2H

6, g) = � 171 + 54 = � 117 kJ


Consider the phase diagram for the binary ideal liquid mixture of '1' and '2' shown in figure and answer thefollowing questions.

33. 'P' represents a point at which the mixture is present in :(A) entirely liquid phase(B) entirely vapour phase(C) both vapour as well as liquid in equilbrium with each other(D) cannot be predicted from the graph

Ans. (A)

34. Which pair of points represent different phases in equilbrium with each other :(A) P and T (B) S and M (C) A and S (D) A and M

Ans. (D)



Delocalisation of electrons take place in conjugated system involving carbon atoms. Delocalisation may alsooccur in a conjugated system involving carbon atoms and atoms other than the carbon. There are examples inwhich orbital and p orbital (vacant or half- filled or filled) overlap. Thus delocalisation are of the following types:(i) delocalisation by overlap.(ii) delocalisation by p overlap.Delocalisation makes system stable. more is the number of resonating structures more is the stability of thesystem .

35. In which of the following compounds delocalisation is not possible :

(A) 1, 4-pentadiene (B) 1,3-butadiene (C) 1,3,5-hexatriene (D) benzeneAns. (A)Sol. H

2C = CH�CH�CH=CH

2 (No conjugation).

36. Which of the following anion is most stable ?

(A) (B)

O = C � O

NO2

(C)

O = C � O

NO2

(D)

O = C � O

NO2

Ans. (B)Sol. � I , � m of �NO

2 group.



37. A non volatile, non electrolyte solute CnH

2nO

n of 2.4 g is dissolved in 100 g water so its boiling point get elevated

upto 100.1ºC (kb of water = 0.5 K-kg/mole).What is the value of 'n'.

Ans. 4Sol. T

B = k

b . m

0.1 = 0.5 × 100m4.2

× 1000

m = 120 g = 12n + 2n + 16n = 30nn = 4

38. How many of the following properties are intensive properties.Density, Temperature, Total heat capacity, Boiling point, Molarity, Enthalpy, Volume, Pressure.

Ans. 5Sol. Only Density, Temperature, Boiling point, Molarity and Pressure are intensive properties (independent of the

size of the system). Rest all are extensive properties.


39. Two mole of an ideal diatomic gas undergoes expansion along a straight line on PV curve from initial state A (3L,8 atm) to final state B (7.5L, 2 atm). Calculate the magnitude of q for the above process in L atm.

Ans. 0

Sol.

WAB

= � Area under curve

= � 21

( 8+2) (7.5 � 3)

= � 22.5 L atm

UAB

= nCVT = 1

2R5

[TB � TT

A]

= 1 2R5

R138

R15.72

= � 22.5 L atm

qAB

= U � W = � 22.5 + 22.5 = 0.

40. Find the total number of -electrons taking part in resonance in the given structure of carbocation?

Ans. 8Sol. One bond has two 2-electrons.

41. How many groups (each attached with benzene ring) show + M effect?

Ans. 5


Sol. have + M group.

42. In how many of the following cases, the negative charge is delocalised?

Ans. 4

Sol. � ve charge is delocalised in each of the

PART-III (MATHEMATICS)


This section contains 5 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of whichONLY ONE is correct.

43. Given that 9 is a root of the equation log(x2 + 15a2) � log

(a � 2) = log

2�axa8 The other root is

(A) 15 (B) 9 (C) 5 (D) 9/5Ans. (A)

Sol. Given that 9 is a root of the equation log(x2

+ 15a2) � log

(a � 2) = log

2�aax8

a > 2

Putting x = 9 , log

2�aa1581 2

= log 2�a

a72

5a2 � 24a + 27 = 0 i.e., a = 3 or 59

a > 2 a = 3. log

(x2 + 135) = log

(24x)

i.e., x2 � 24x + 135 = 0.

Roots are 9 and 15.The other root is 15.


44. If [sin-1 cos�1 sin�1 tan�1 x] = 1, where [.] denotes the greatest integer function, then x belongs to the interval(A) [tan sin cos 1, tan sin cos sin 1] (B) (tan sin cos 1, tan sin cos sin 1)(C) [�1, 1] (D) [sin cos tan 1, sin cos sin tan 1]

Ans. (A)Sol. [sin�1 cos�1 sin�1 tan�1 x] = 1

1 sin�1 cos�1 sin�1 tan�1 x 2

sin 1 cos�1 sin�1 tan�1 x sin 2

= 1

cos 1 sin�1 (tan�1 x) cos (sin 1) sin (cos 1) tan�1 x sin {cos(sin 1)} tan sin cos 1 x tan sin cos sin 1

45. Let f(x), g(x), h(x) be quadratic polynomials with real coefficients and coefficient of x2 being positive.Further, zeroes of each of these polynomials are real and distinct. If every two of these polynomials have acommon zero then roots of the equation f(x) + g(x) + h(x) = 0 are(A) real and distinct (B) real and equal(C) imaginary (D) none of these

Ans. (A)Sol. Let f(x) = a

1(x � )(x � )

g(x) = a2 (x � )(x � )

h(x) = a3 (x � )(x � )

where a1, a

2, a

3 are positive

Let f(x) + g(x) + h(x) = F(x) F() = a

2 ( � )( � )

F() = a3 ( � )( � )

F() = a1 ( � )( � )

F() F() F() = � a1a

2a

3( � )2 ( � )2 ( � )2 = negative

roots of F(x) = 0 are real and distinct.

46. If cos 28º + sin 28º = k3, then cos17º is equal to

(A) 2

k3

(B) �2

k3

(C) 2

k3

(D) none of these

Ans. (A)Sol. cos 17º = cos (45º � 28º) = cos 45º cos28º + sin45º sin28º

= 2

º28sinº28cos =

2

k3

.

47. The number of values of x satisfying

2x

+

4x

+

6x

= 12

x11, x [0, 100], where [ . ] represents greatest

integer function, are(A) 8 (B) 9 (C) 10 (D) none of these

Ans. (B)

Sol.

2x

+

4x

+

6x

= 12

x11 = integer number

Let x = 12k , then

2

x +

4

x +

6

x = 6k + 3k + 2k = 11k. Also

12x11

= 11k1k

statement is true for all integral value of k. x must be multiple of 12. x = 0, 12, 24, 36, 48, 60, 72, 84, 96 which are 9 in number


SECTION - IIMultiple Correct Answer Type

This section contains 4 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of whichONE OR MORE is/are correct.

48. If the quadratic equation ax2 + 2bx � 4 = 0, where a, b, c R, does not have two real & distinct roots, then(A) a � b > 1 (B) 2b � a �4 (C) 2b � a �4 (D) a + b 1

Ans. (CD)Sol. Let f(x) = ax2 + 2bx � 4

f(x) = 0 does not have two real & distinct roots and f(0) = �4

f(x) 0 x Rfurther f(�2) = 4a � 4b � 4 0 a � b 1f(�1) = a � 2b � 4 0 2b � a �4

f(2) = 4a + 4b � 4 0 a + b 1

49. Consider the system of equations ax + by = 1, cx + dy = 0. The system has no solution if(A) ad � bc 0 (B) ad � bc = 0, c 0(C) ad � bc = 0, d 0 (D) ad � bc = 0, c2 + d2 = 0

Ans. (BC)Sol. The system has no solution if

ac

= bd

10

i.e.if ad � bc = 0 and c 0, d 0

50. If x = sec tan & y = cosec + cot then

(A) x = y

y

1

1(B) y =

1

1

x

x(C) x =

y

y

1

1(D) xy + x y + 1 = 0

Ans. (BCD)Sol. x = sec tan & y = cosec + cot

y =

sincos1

= )2/sin(2)2/cos(2

= cot

2

x�1x1

=

tansec�1tan�sec1

=

cos/)sin1�(coscos/)sin�1(cos

=

sin1�cossin�1cos

=

sin2

sin2�

sin�2

cos2

2

2

=

2sin2�

2cos.

2sin2

2cos

2sin2�

2cos2

2

2

=

2sin2

2cos2

2sin�

2cos

2sin�

2cos

= cot 2

= cosec + cot = y

also y � xy = x + 1

y � 1 = x(y + 1)

x = 1y1�y


51. If graph of y = f(x) is as shown in figure

then which of the following options is/are correct ?

(A) Graph of y = f(�|x|) is (B) Graph of y = f(|x|) is

(C) Graph of y = |f(|x|)| is (D) Graph of |y| = f(x) is

Ans. (ACD)

Sol. Obviously by graphical transformation.


This section contains 3 paragraphs. Based upon each paragraph, there are 2 questions. Each question has 4choices (A), (B), (C) and (D), out of which ONLY ONE is correct.


Let the domain and range of inverse circular functions be defined as followsDomain Range

sin�1x [�1, 1]

2�,

23

�

cos�1x [�1, 1] [, 2]

tan�1x R

2�,

23

�

cot�1x R (, 2)

cosec�1x (�, �1] [1, )

2�,

23

� � {�}

52. If sin�1x < 6

5�

, then solution set of x is

(A)

1,

2

1(B)

1,

2

1� (C)

21

,21

(D)

21

�,1�

Ans. (B)

Sol.23

sin�1x < 6

5�

sin

6

5 < x sin

2

3

21

< x 1


53. If f(x) = cos�1x + tan�1x + cot�1x + cosec�1x, then f(1) is equal to(A) (B) 2 (C) 3 (D) �

Ans. (A)Sol. f(1) = cos�1 1 + tan�1 1 + cot�1 1 + cosec�1 1

= 2 � 4

3 +

45�

23

=


Let ax2 + a2x + 2 = 0 be a quadratic equation, a R and S be the set of values of a for which roots ofthis equation are imaginary or equal, then answer the following questions.

54. Number of integral values of 'a' in set S are(A) 1 (B) 2 (C) 3 (D) 4

Ans. (B)Sol. ax2 + a2x + 2 = 0 has imaginary or equal roots if

(a2)2 � 4(2) (a) 0i.e a(a3 � 8) 0i.e a(a � 2) (a2 + 4 + 2a) 0i.e a (0, 2] (as a 0) S (0, 2]number of integral values in set S is 2

55. If and are respectively sum and product of roots of the given equation, then � is(A) a prime number (B) an odd integer(C) an irrational number (D) dependent on value of a

Ans. (A)

Sol. � = �

a2

aa2

= 2

prime number a R


Consider the equation ||x2 � 2| � 2| + 2 = , where R

56. If given equation has four real solutions, then(A) {2} (B) (2, 4) (C) {4} (D) {2, 4}

Ans. (C)

57. If given equation has two real solutions, then(A) {2} (B) (2, 4) (C) (4, ) (D) {4}

Ans. (C)Sol. Graph of y = ||x2 � 2| � 2| + 2

(i) for 4 solution : = 4(ii) for 2 solution : > 4




58. If the value of sin18

sin183

sin185

sin187

sin189

sin1811

sin18

13sin

1815

sin 18

17 is then find the value

of 512Ans. 2

Sol. sin18

sin183

sin185

sin187

sin189

sin

187

� sin

185

� sin

183

� sin

18�

=

2

187

sin185

sin183

sin18

sin

189

sin =

2

9cos

92

cos9

3cos

94

cos

=

2

94

cos92

cos9

cos41

=

2

3

3

9sin2

92

sin

41

= 41

2

81

= 41

× 641

= 256

1

59. If the number of values of x satisfying the equation 2x + 3 [x] � 4 {�x} = 4 are , then find the value of 2 (where[x] and {x} denote integral and fractional part of x respectively)

Ans. 2Sol. 2x + 3 [x] � 4 {�x} = 4

2x + 3 [x] � 4 (�x � [�x]) = 4

2x + 3 [x] + 4x + 4 [�x] = 4

Case-I : If x 2x + 3x + 4x � 4x = 4

5x = 4 x 54

Case-II : If x [�x] = � [x] � 1

2x + 3 [x] + 4x � 4 [x] � 4 = 4

6x � [x] = 8

6x = [x] + 8 ........(i) 6[x] + 6{x} = [x] + 8 6{x} = 8 � 5 [x]

{x} = 6

]x[58

0 6

]x[58 < 1

0 8 � 5[x] 6

� 8 � 5 [x] < �2 52

< [x] < 58

[x] = 1

by (i) 6x = 9 x = 23

= 1


60. If , are roots of equation x2 � 2x + 3 = 0 then find the value of 3 + 3 � 32 � 2 + 5+ Ans. 0Sol. + = 2, = 3

3 + 3 � 32 � 2 + 5+ =( + )3� 3 (+ ) � (2+ 2) � 22 + 4+ ( + ) = 0

61. Find the number of ordered pairs (x, y) satisfying the system of equations

xy20

= y�xyx ,

y5x16

= y�x�yx .

Ans. 1

Sol.x

y20 = yx + y�x ....... (i)

xy20

=

y5x16

y2

y�x�yx

yx�yx

=

x4

y5y2 [using eqn (ii)]

y = 4 x = 5one order pair (5, 4)

62. If the complete solution set of the equation 21x10�x

12x8�x2

2

= �

21x10�x

)12x8�x(2

2

is [, ) [, ) then the

value of + � ( + ) isAns. 0

Sol. The equation 21x10�x

12x8�x2

2

= �

21x10�x

)12x8�x(2

2

21x10�x

12x8�x2

2

0

x [2, 3) [6, 7)

63. If 4x � 3x � 1/2 = 3x + 1/2 � 22x � 1, then value of 2x is equal toAns. 3Sol. 4x � 3x � 1/2 = 3x + 1/2 � 22x � 1

4x + 22x � 1 = 3x + 1/2 + 3x � 1/2

4x + 2

4x

= 3x 3 + 3

3x

4x

21

1 = 3x

3

13

x

3

=

32

× 3

4 =

33

8 =

3

2/13

2

=

2/3

34

x = 3/2 2x = 3

Documents

VIJETA (JP) IPT-1 Solution Booklet English 22-04-2012[1]