Jee Main Physics Solutions 2015

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[ 1 ]IITians TAPASYA...IITians creating IITians

SOLUTION OF JEE MAIN. 2015

PART - B [PHYSICS]31. In the circuit shown, the current in the 1 resistor is :

(1) 0.13 A, from P to Q(2) 1.3 A, from P to Q(3) 0A(4) 0.13 A, from Q to P

Ans. (4) For calculation of current, circuit can be drawn as :

Apply Nodal methodx 0 x 6 x 9 0

1 3 5

6

6

3 x 5 9

9

0

13x

23

Current in 1 =x 0 3 0.13

1 23

Current is 0.13 A from Q to P.32. Distance of the centre of mass of a solid uniform cone from its vertex is z0. If the radius

of its base is R and its height is h then z0 is equal to :

(1)23h

8R(2)

2h4R

(3)3h4 (4)

5h8

Ans. (3) 223Mdm r dxR H

x

dx

Here Ycm =H

0

1 dmxM

IITians TAPASYA...IITians creating IITians



[ 2 ]IITians TAPASYA...IITians creating IITians2H

20

1 3M Rx dx xM R H H

H3

30

3 x dxH 3H4

33. Match List - I (Fundamental Experiment) with List - II (its conclusion) and select the correct optionfrom the choices given below the list :

List - I List - II(1) Franck-Hertz Experiment (i) Particle nature of light(B) Photo-electric experiment (ii) Discrete energy levels of atom(C) Davision - Germer Experiment (iii) Wave nature of electron

(iv) Structure of atom(1) (A) -(iv), (B) -(iii), (C) -(ii) (2) (A) -(i), (B) -(iv), (C) -(iii)(3) (A) -(ii), (B) -(iv), (C) -(iii) (4) (A) -(ii), (B) -(i), (C) -(iii)

Ans. (4)(1) Franck-Hertz experiment :- In 1914, Franck-Hertz experiment demonstrated the existence of

excited states in mercury atoms.(2) Photo-electric effect is explained by particle nature of light.(3) Davisson-Germer experiment confirmed the De-broglie hypothesis which says that particles of matter such as electron has wave nature also.

34. The period of oscillation of a simple pendulum is T =L2g

. Measured value of L is 20.0 cm

known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s usinga wrist watch of 1s resolution. The accuracy in the determination of g is :(1) 5% (2) 2% (3) 3% (4) 1%

Ans. (3)g L T2

g L T

L 0.1cm, L 20cm

t 1 90T 0.01s, T 0.9sn 100 100

g 0.1 0.012 0.005 0.222 0.027g 20 0.9

g% Accuracy 100% 2.7% 3%g

35. A red Led emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the lightat a distance of 1 m from the diode is :(1) 7.75 V/m (2) 1.73 V/m (3) 2.45 V/m (4) 5.48 V/m

Ans. (3) Intensity I 20 01 E C2



[ 3 ]IITians TAPASYA...IITians creating IITians2

0 02P 1 E C

4 r 2

(where P = power of LED)

920 2 8 2

0

2P 2 0.1 9 10E 64 cr 3 10 (1)

0E 6 2.45 v /m

36. In the given circuit, charge Q2 on the 2 F capacitor changes asC is varied from 1F to 3F. Q2 as a function of C is givenproperly by : (figure are drawn shcematically and are not to scale)

(1) (2)

(3) (4)

Ans. (3) Charge in series are same.Let potential on c is V1 & potential on 3F is V2 .CV1 = 3V2 ...(1)

C

E

3FV1 + V2 = E ...(2)From (1) and (2)

2CEV

(C 3)

Now, Q2 = 2V2

22CEQ

(C 3)



[ 4 ]IITians TAPASYA...IITians creating IITians37. Two long current carrying thin wires, both with current I, are held by insulating threads of length L

and and are in equiblibrium as shown in the figure, with threads making an angle with thevertical. If wire have mass per unit length then the value of I is :(g = gravitational acceleration)

(1)0

gL tan (2)

0

gLsincos

(3)0

gL2sinu cos

(3)0

gL2 tanu

Ans. (3) Making free body diagram of unit length of a wire

T

g

20I

2 (2Lsin )

Vertical equilibrium : Tcos g ....(1)

Horizontal equilibrium :2

0ITsin4 Lsin

....(2)

Using equation (1) and (2), we get

0

LgI 2sincos

38. A particle of mass m moving in the x direction with speed 2 is hit by another particle of mass 2mmoving in the y direction with speed . If the collision is perfectly inelastic, the percentage lossin the energy during the collision is close to :(1) 62% (2) 44% (3) 50% (4) 56%

Ans. (4) Initial Momentum of System iP = 2 2(2mv) (2mv) 2 2mv Final Momentum of System = PfConservation of Linear Momentum: iP = Pf

1 12 22 2 mV 3m V V V

3

Initial kinetic energy Ki =2 2 21 1m(2V) 2mV 3mV

2 2

Final kinetic energy Kf =2 21 8 4(3m) V mV

2 9 3

2 2 2i f

4 5K K K 3mV mV mV3 3

2

2i

K 5 / 3 mV 5% loss 100 56%K 3mV 9



[ 5 ]IITians TAPASYA...IITians creating IITians39. Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively. These are

being pressed against a wall by a force F as shown. If the coefficient of friction between theblocks is 0.1 and between block B and the wall is 0.15,the frictional force applied by the wall on block B is :(1) 150 N(2) 100 N(3) 80 N(4) 120 N

Ans. (4) FBD of A and B

FF F F

f1

f120 N

A B

100 N

f2

For A, f1 = 20 NFor B, f1 + 100 = f2 = 120 N

40. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adia-batic expansion, the average time of collision between molecules increases as Vq, where V is

the volume of the gas. The value of q is :p

v

CC

(1)1

2

(2)3 5

6

(3)3 5

6

(4)1

2

Ans. (4) Average time of collision

2

kT Tt t & TP2 P

T TtPTP

V T Vt tT T

For adiabatic process 1TV cons tan t 1t V V

1 112 2t V t V

Given qt V , Hence, q =1

2



[ 6 ]IITians TAPASYA...IITians creating IITians41. A rectangular loop of sides 10cm and 5cm carrying a current I of 12 A is placed in different

orientations a shown in the figures below :

(A) (B)

(C) (D)

If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium?(1) (b) and (c), respectively (2) (a) and (b), respectively(3) (a) and (c), respectively (4) (b) and (d), respectively

Ans. (4) Magnetic moment M

& Magnetic field B

(i) If M

|| B

, then loop will be in stable equilibrium.(ii) If M

antiparallel to B

, then loop will be in un stable equilibrium.

Hence, (b) shows stable equilibrium(d) shows unstable equilibrium

42. Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be

considered as an ideal gas of photons with internal energy per unit volume u =UV T4 and

pressure p =1 U3 V

. If the shell now undergoes an adiabatic expansion the relation between T

and R is :

(1) 31T

R (2) RT e (3) 3RT e (4)

1TR

Ans. (4) 4 4U 1 Uu T & P P T ..........(i)V 3 V

from Ideal gas equation :TPV nRT P ..............(ii)V

from equation (i) & (ii)

31/ 3

1 1T TV V

Volume 3 1/ 3V R V R

Hence,1TR



[ 7 ]IITians TAPASYA...IITians creating IITians43. As an electron makes a transition form an excited state to the ground state of a hydrogen- like

atom/ion :(1) kinetic energy and total energy decrease but potential energy increases(2) its kinetic energy increases but potential energy and total energy decreases.(3) kinetic energy, potential energy and total energy decreases.(4) kinetic energy decreases, potential energy increases but total energy remains same.

Ans. (2) VelocityZVn , As electron goes to lower orbit, the velocity of electron increases.

hence, kinetic energy increases Total energy decreases because electron emits photon. Total energy (T.E) = Potential energy (P.E) + kinetic energy (K.E)

If K.E & T.E P.E 44. On a hot summer night, the refractive index of air is smallest near the ground and increases with

height from the ground. When a light beam is directed horizontally, the Huygens principle leadsu sto conclude that as it travels, the light beam :(1) bends upwards (2) becomes narrower(3) goes horizontally without any deflection (4) bends downwards.

Ans. (1)If a wavefront is moving in horizontal direction then upward rays will move slower then lower rays

hence if rays are going in forward direction then wavefront will have a tendency to bend backward and the light beam will bend upwards.45. From a solid sphere of mass M and radius R, a spherical portion of radius R/2 is removed, as

shown in the figure. Taking gravitational potential V = 0 at r , the potential at the centre of thecavity thus formed is :(G = gravitational constant)

(1)2GMR

(2)GM2R

(3)GMR

(4)2GM3R

Ans. (3) Potential inside a solid sphere is given by

2 22V G (3R r )3

In the cavity we can assume that both and density material is filledV = V solid sphere + Vcavity

2 222 R 2 RV G 3R G 3 0

3 2 3 2

+ &

24V R G3

By Putting3

M4 R3

, We get GMVR

46. Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material



[ 8 ]IITians TAPASYA...IITians creating IITiansof the prism is , a ray, incident at an angle , on the face AB would get transmitted through theface AC of the prism provided.

(1) 1 1 1cos sin A sin

(2) 1 1 1sin sin A sin

(3) 1 1 1sin sin A sin

(4) 1 1 1cos sin A sin

Ans. (2) If 2 Cr (critical angle)1

21r sin ...........(i)

Then ray will always transmit through face AC

At face AB : 11 11sin sinr r sin sin

r1 r2

A

B C

We know for prism r 1 + r2 = A

r2 = A r1 = A 1 1sin sin

................ (ii)

from equation (i) and (ii)

1 11 1A sin sin sin

1 11 1A sin sin sin

1 1 1sin A sin sin

1 1 1sin sin A sin

47. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with intial speed of10 m/s and 40 m/s respectively. Which of the following graph best represents the time variationof relative position of the second stone with respect to the first?

(1) (2)




(3) (4)

Ans. (4)Initially both the particles will be in air and moving under gravity. Their relative acceleration will bezero. So relative separation Vs time will be straight line.When one particle reaches the ground then relative acceleration will be g.Now relative separation vs time is parabolic with concave downwards as relative speed shouldincrease with time.

48. For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy(PE) against its displacement d. Which one of the following represents these correctly ?(graphs are schematic and not drawn to scale)

(1) (2)

(3) (4)

Ans. (3) At Extreme positions K.E = 0At Mean position K.E is maximumAt Extreme positions P.E is maximumAt Mean position P.E is minimumSince, T.E. in S.H.M remains conserved.d = 0 Mean positiond = A Extreme positions (A Amplitude)

49. A train is moving on a straight track with speed 20 ms1. It is blowing its whistle at the frequencyof 1000 Hz. The percentage change in the frequency heard by a person standing near the trackas th e train passes is (speed of sound = 320 ms1) close(1) 24% (2) 6% (3) 12% (4) 18%

Ans. (3) For approaching case :

app oVf f

V 20

For moving away case :

away oVf f

V 20

oapp away o 2

40f V1 1f f f f VV 20 V 20 V 400



[ 10 ]IITians TAPASYA...IITians creating IITians% change =

2of 40 x320 x100 40 x320 x100x100% 12.54% 12%

f 102000320 400

50. An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged toQ0 and then connected to the L and R as shown below :

If a student plots graphs of the square of maximum charge 2MaxQ on the capacitor with time (t)for two different values L1 and L2 (L1 > L2) of L then which of the following represents this graphcorrectly ? (plots are schematic and not drawn to scale)

(1) (2)

(3) (4)

Sol. (2) Damping is more, if inductance is less

51. A solid body of constant heat capacity 1 J/oC is being heated by keeping it in contact with reservoirsin two ways :(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same

amount of heat.(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same

amount of heat.In both the cases body is brought from initial temperature 100oC to final temperature 200oC.Entropy change of the body in two cases respectively is :

(1) l l2 n2,8 n2 (2) l ln2,4 n2 (3) l ln2, n2 (4) l ln2, 2 n2

Ans. (3)dq CdTdsT T

, temp should be given in kelvin

For both the case : 200100

s 1 lnT | 200100s 1 lnT | ln 252. An inductor (L = 0.03H) and resistor (R = 0.15 k ) are connected in series to a battery of 15V



[ 11 ]IITians TAPASYA...IITians creating IITiansEMF in a circuit shown below. The key K1 has been kept closed for a long time. Then at t = 0, K1is opened and key K2 is closed simultaneously. At t = 1ms, the current in the circuit will be :( 5e 150 )

(1) 0.67 mA(2) 100 mA(3) 67 mA(4) 6.7 mA

Ans. (1) Current at t =0

o 315I 0.1 A

0.15 x 10

Now current will decrease with time as per equation3 30.15x10 x10

Rt /L 50.03o

0.1I I e 0.1 e 0.1 e 0.67 mA150

53. A uniformly charged solid sphere of radius R has potential V0(measured with respect to ) o n

its surface. For this sphere the equipotential surfaces with potentials 0 0 0 03V 5V 3V V, , and2 4 4 4

have radius R1, R2, R3 and R4 respectively. Then(A) 2R < R4 (B) R1 = 0 and R2 > (R4 R3)

(C) R1 0 and (R2 R1) > (R4 R3) (D) R1 = 0 and R2 < (R4 R3)

Sol. (1, 4)

2o

o 2VKQ 3 rV , V

R 2 R for r < R and 0V RV , r R

r

0centre 1

3VV R 04

02

5V RV R4 2

03

3V 4RV R4 3

for 0 4VV R 4R4

Hence, R4 > 2R and 4 3 2R R R

54. A long cylindrical shell carries positive surface charge in the upper half and negative surface charge in the lower half. The electric field lines around the cylinder will look like figure given in :




(1) (2) (3) (4)

Sol. (2)(i) Electric field line curves should be smooth(ii) Electric field lines originates from positive charge & terminates at negative charge(iii) Tangent at any point on curve shows direction of resultant electric field.(iv) Preference of termination should be at negative charge but in option (4) no such termination

is shown.55. Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the

minimum separation between two objects that human eye car resolve at 500 nm wavelength is :(1) 300 m (2) 1 m (3) 30 m (4) 100 m

Sol. (3)

object 1

object 225 cm

For just resolve the images

0.61a

where a radius of pupilFrom figure

d 0.61 d 3025cm a

56. A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. Thefrequencies of the resultant signal is/are :(1) 2000 kHz and 1995 kHz (2) 2 MHz only(3) 2005 kHz and 1995 kHz (4) 2005 kHz, 2000 kHz and 1995 kHz

Sol. (4) Amplitude modulated wave carries three frequencies = WC W, WC, WC + WWC - frequency of carrier waveW - frequency of signalWC = 2 MHz or 2000 KHzW = 5 KHzHence, three frequencies in resultant signal are 2005 KHz, 2000 KHz, 1995 KHz

57. Two coaxial solenoids of different radii carry current I in the same direction. Let 1F

be the magneticforce on the inner solenoid due to the outer one and 2F

be the magnetic force on the outer

solenoid due to the inner one. Then :

(1) 1F

is radially outwards and 2F

= 0



[ 13 ]IITians TAPASYA...IITians creating IITians(2) 1F

= 2F

= 0

(3) 1F

is radially inwards and 2F

is radially outwards

(4) 1F

is radially inwards and 2F

= 0

Sol. (2) Inner solenoid is kept in uniform magnetic field of outer solenoid so force on all currectloops is zeroInner solenoid produce zero magnetic field outside. So force on outer solenoid is also zero.

58. A pendulum made of a uniform wire of cross sectional area A has time period T. When an additionalmass M is added to its bob, the time period changes to TM. If the Youngs modulus of the material

of the wire is Y then1Y is equal to :

(g = gravitational acceleration)

(1)2

M

T A1T Mg

(2)2

MT A1T Mg

(3)

2MT Mg1

T A

(4)2

MT A1T Mg

Sol. (2)

M

lT 2 ...(1)g

Mg lY ....(2)A l

l lT 2 ....(3)g

59. From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Momentof inertia of cube about an axis passing through its center and perpendicular to one of its facesis :

(1)24MR

3 3 (2)2MR

32 2 (3)2MR

16 2 (4)24MR

9 3

Sol. (4) From symmetry it is obvious that for maximum volume

3a 2R

Mass of cube = 33

2cube

M 2Ma4 3R3

M aI6

60. When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons



[ 14 ]IITians TAPASYA...IITians creating IITiansis 2.5 x 104 ms1. If the electron density in the wire is 8 x 1028 m3, the resistivity of the material isclose to :

(1) 1.6 x 105 m (2) 1.6 x 108 m (3) 1.6 x 107 m (4) 1.6 x 106 m .

Sol. (1) l

l

d

5

d

II neAV , V IRA

V 1.56 10neV

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Jee Main Physics Solutions 2015