Jee Main Physics Solutions 2015

IITian’s TAPASYA

[ 1 ]

. . .I .I I Ti Ti a n s c r e a t i n g I I T i a n s

I I T i a n ’ s T A PA S Y A . . . I ITians c reatin g I I Tians

SOLU SOLU TION OF JEE MAIN. 2015 PART - B [PHYSICS] 31. 31.

In the the circ circui uitt sho show wn, the the curr curren entt in in the 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. Ans.

(4) For calculation of current, circuit can be drawn as : x0 Apply Nodal method 1



x

Current in 1 =

32. 32.



x9 5

–6

0

3

3



3 23

5

1

6

23

x0 1

x

9

9

0

  0.13

Current is 0.13 0.1 3 A from Q to P. P. Distan Distance ce of the the cent centre re of of mass mass of a solid solid unifor uniform m cone cone from from its its vert vertex ex is is z0. If the radius of its base is R and its height is h then z0 is equal to : 3h2 (1) 8R

Ans. Ans.



x6 3

(3) dm 

h2 (2) 4R

(3)

3M   r 2 dx 2 R H H

1 dm x Here Ycm = M0

3h 4

(4)

5h 8

x dx



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[ 2 ]

. . .I I Ti a n s c r e a t i n g I I T i a n s

 1  3M  Rx    2    dx  x M 0  R H  H   2

H



H

3  3 x 3dx  3H H 0 4



33.

Ans.

34.

Match List - I (Fundamental Experiment) with List - II (its conclusion) and select the correct option from 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) (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. The period of oscillation of a simple pendulum is T = 2

L . Measured value of L is 20.0 cm g

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

(3)

(2) 2%

g g



L L

2

g g



t n



T

1 100

L

 20 cm

 0.01s,

0.1 0.01  2 20 0.9

% Accuracy  35.

(4) 1%

T

L  0.1cm,

T 

(3) 3%

g g

T

90 100

 0.9 s

 0.005  0.222  0.027

 100%  2.7%  3%

A red Led emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is : (1) 7.75 V/m

Ans.

(3) Intensity

(2) 1.73 V/m I

(3) 2.45 V/m

(4) 5.48 V/m

1 2

 0E02C

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[ 3 ]


P 4 r 2

2P 40cr 2



(where P = power of LED) 2  0.1 9  109 3  108  (1)2

E



E0

 6  2.45 v / m

2 0

36.

1 2

 0E02C

6

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

Ans.

(1)

(2)

(3)

(4)

(3)

Charge in series are same. Let potential on ‘c’ is V1 & potential on 3  F is V2 . CV1 = 3V2 ...(1) V1 + V2 = E ...(2) From (1) and (2) V2



3 F

C

CE (C  3)

E

Now, Q2 = 2V2 Q2



2CE (C  3)



IITian’s TAPASYA

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37.

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 the vertical. If wire have mass  per unit length then the value of I is : (g = gravitational acceleration)

gL tan  (1) 0 (3) 2sin  Ans.

gL 0 cos

(2) sin 

gL u0 cos 

(3) 2

gL u0

tan 

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

 0I2 2(2L sin )

g Vertical equilibrium :

T cos   g

....(1)

 I2 0 Horizontal equilibrium : Tsin   4L sin 

....(2)

Using equation (1) and (2), we get I  2 sin  38.

Lg  0 cos 

A particle of mass m moving in the x direction with speed 2  is hit by another particle of mass 2m moving in the y direction with speed

. If the collision is perfectly inelastic, the percentage loss

in the energy during the collision is close to : (1) 62% Ans.

(2) 44%

(4) Initial Momentum of System Pi =

(3) 50% (2mv)2  (2mv)2

(4) 56%

 2 2mv

Final Momentum of System = Pf Conservation of Linear Momentum: Pi = Pf 2 2 mV

 3m V1  V1 

2 2 V 3

1 1 m(2V)2  2mV 2  3mV 2 2 2 1 8 2 4 2 Final kinetic energy Kf = (3m) V  mV 2 9 3 4 5 K  K i  K f  3mV 2  mV2  mV2 3 3

Initial kinetic energy Ki =

% loss 

K Ki



5 / 3 mV 2 3mV 2

5 9

  100  56%



IITian’s TAPASYA

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39.

Ans.

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 coeff icient of friction between the blocks 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 (4) FBD of A and B

f 2

f 1 F

A

F

F

20 N

40.

B

F

f 1 100 N

For A, f 1 = 20 N For B, f 1 + 100 = f 2 = 120 N Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as Vq, where V is

 Cp   the volume of the gas. The value of q is :     Cv  3  5  1

(1) Ans.

(2)

2 (4) Average time of collision

kT

t



t



T

t



V T T

2 P 2



TP

 t

6

T & P

(3)

3  5 6

(4)

 1 2

 T

T P

 t 

V T

For adiabatic process TV  1  cons tan t t

t Given t

 V V 1

 V 

1

q

1 2

1

 t V

V , Hence, q =

2

 1 2



IITian’s TAPASYA

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41.

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 orienta tions 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 U  T4 and considered as an ideal gas of photons with internal energy per unit volume u = V pressure p =

1 U   . If the shell now undergoes an adiabatic expansion the rel ation between T 3V

and R is :

Ans.

(1) T 

1 R3

(4) u 

U V

 T4 & P 

1U 3V

(4) T 

(3) T  e 3R

(2) T  e R

1 R

 P  T 4 ..........(i)

from Ideal gas equation :

PV

 nRT  P 

T V

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

from equation (i) & (ii) T3



1 V

 T

Volume V Hence,

1 V1/ 3

 R3  V1/ 3  R T



1 R



IITian’s TAPASYA

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43.

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 potent ial 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) Velocity V 

 

Z

, As electron goes to lower orbit, the velocity of electron increases. n 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. W hen a light beam is directed horizontally, the Huygen’s principle leads u 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 the cavity thus formed is :(G = gravitational constant) (1) (3)

2GM

(2)

R

GM

R (3) Potential inside a solid sphere is given by

Ans.

3R

3

=

  and   density material is filled

V

 2  R 2  2 V   G  3R      G  3  2   3  2

  R 2   3    0    2 

+ & –

4

  R2G 3

By Putting

46.

2GM

2

In the cavity we can assume that both + Vcavity solid sphere

V

2R

  G (3R2  r 2 )

V

V

(4)

GM



M

,

We get V 

GM

4 R R3 3 Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material Head Office : Pushpanjali Place, Boring Road Crossing, Opp. Alankar Place, Patna-1 Branch Office : Rukanpura Flyover, Near Sri Sai Yamaha, Bailey Road, Patna- 14 Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777


IITian’s TAPASYA

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of the prism is  , a ray, incident at an angle  , on the face AB would get transmitted through the face AC of the prism provided.

   1     cos1  sin  A  sin1               1   (2)   sin1  sin  A  sin1               1   (3)   sin1  sin  A  sin1               1   (4)   cos1  sin  A  sin1            (1)

Ans.

(2) If r 2

 1  C (critical angle)  r2  sin1   

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

Then ray will always transmit through face AC

A

1  At face AB : sin    sinr1  r1  sin  sin     1

We know for prism

r

+ r 2 = A

1



1  r 2 = A – r 1 = A – sin  sin   ................ (ii)   1

from equation (i) and (ii)

B

1   1 sin    sin1     

r 1

r 2 C

A  sin1 

 1  1  1   sin sin      

A  sin1 



 1  1     sin    

sin  A  sin1 



  



 1         

sin1   sin  A  sin1 

47.



Two stones are thrown up simultaneously from the edge of a cliff 240 m high with intial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first?

(1)

(2)



IITian’s TAPASYA

[ 9 ]


(3)

Ans.

48.

(4)

(4) Initially both the particles will be in air and moving under gravity. Their relative acceleration will be zero. 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 should increase with time. 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 represent s these correctly ? (graphs are schematic and not drawn to scale)

Ans.

(1)

(2)

(3)

(4)

(3)

At Extreme positions K.E = 0 At Mean position K.E is maximum At Extreme positions P.E is maximum At Mean position P.E is minimum Since, T.E. in S.H.M remains conserved. d = 0  Mean position

49.

d =  A  Extreme positions (A  Amplitude) A train is moving on a straight track with speed 20 ms –1. It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near t he track as th e train passes is (speed of sound = 320 ms –1) close (1) 24%

Ans.

(2) 6%

(3) 12%

(4) 18%

(3) For approaching case : fapp

V   f o    V  20 

For moving away case : faway

 V   f o    V  20 

40fo V 1 1  f  fapp  faway  fo V     V 2  400  V  20 V  20  Head Office : Pushpanjali Place, Boring Road Crossing, Opp. Alankar Place, Patna-1 Branch Office : Rukanpura Flyover, Near Sri Sai Yamaha, Bailey Road, Patna- 14 Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777


IITian’s TAPASYA

[ 10 ]


% change = 50.

f fo

x100% 

40 x 320 x100

  320

2

 400 



40 x 320 x100 102000

 12.54%  12%

An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q0 and then connected to the L and R as shown below :

If a student plots graphs of the square of maximum charge  QMax  on the capacitor with time (t) 2

for two different values L1 and L2 (L1 > L2) of L then which of the following represents this graph correctly ? (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 reservoirs in 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 100 oC to final temperature 200oC. Entropy change of the body in two cases respectively is : (1) 2l n2,8l n 2

Ans.

(3)



ds 

dq T



(2) l n 2, 4l n 2

l n 2, l n 2

(4)

l n 2,

2l n 2

CdT , temp should be given in kelvin T

For both the case : 52.

(3)

200 200 s  1 ln T |100 s  1 ln T |100  ln  2 

An inductor (L = 0.03H) and resistor (R = 0.15 k ) are connected in series to a battery of 15V Head Office : Pushpanjali Place, Boring Road Crossing, Opp. Alankar Place, Patna-1 Branch Office : Rukanpura Flyover, Near Sri Sai Yamaha, Bailey Road, Patna- 14 Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777


IITian’s TAPASYA

[ 11 ]


EMF in a circuit shown below. The key K1 has been kept closed for a long time. Then at t = 0, K1 is opened and key K2 is closed simultaneously. At t = 1ms, the current in the circuit will be : ( e5  150 ) (1) 0.67 mA (2) 100 mA (3) 67 mA (4) 6.7 mA Ans.

(1) Current at t =0 Io



15

 0.1 A

0.15 x 103

Now current will decrease with time as per equation

53.

0.15x103 x103 0.03

0.1  0.67 mA 150 A uniformly charged solid sphere of radius R has potential V0(measured with respect to  ) o n

I  Ioe Rt / L

 0.1 e



 0.1 e5 

its surface. For this sphere the equipotential surfaces with potentials

3V0 5V0 3V0 V , , and 0 2 4 4 4

have radius R1, R2, R3 and R4 respectively. Then (A) 2R < R4 (C) R1 Sol.

(B) R1 = 0 and R2 > (R4 – R3)

 0 and (R2 – R1) > (R4 – R3)

Vo  3  r 2  KQ (1, 4) Vo  , V   for r < R and V R 2  R2  Vcentre

3V0



4

V



5V0

V



3V0 4

4

for V 

V0 4



V0R r

,r

R

 R1  0

 R2 

R

 R3 

4R 3

2

 R 4  4R

Hence, R4 > 2R and R 4  R3 54.

(D) R1 = 0 and R2 < (R4 – R3)

 R2

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 :



IITian’s TAPASYA

[ 12 ]


(1) Sol.

(2)

(3)

(4)

(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

Sol.

(2) 1  m

(3)

(3) 30  m

(4) 100  m

object 1

For just resolve the images





0.61 a



object 2 25 cm

where a radius of pupil From figure d 25 cm 56.

Sol.



0.61 a

 d  30

A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies 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

(4) Amplitude modulated wave carries three frequencies = WC – W, W C, WC + W W C - frequency of carrier wave W - frequency of signal W C = 2 MHz or 2000 KHz W = 5 KHz Hence, 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 F1 be the magnetic 

force on the inner solenoid due to the outer one and F2 be the magnetic force on the outer solenoid due to the inner one. Then : 



(1) F1 is radially outwards and F2 = 0 Head Office : Pushpanjali Place, Boring Road Crossing, Opp. Alankar Place, Patna-1 Branch Office : Rukanpura Flyover, Near Sri Sai Yamaha, Bailey Road, Patna- 14 Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777


IITian’s TAPASYA

[ 13 ]

. . .I I Ti a n s c r e a t i n g I I T i a n s 



(2) F1 = F2 = 0 



(3) F1 is radially inwards and F2 is radially outwards 



(4) F1 is radially inwards and F2 = 0 Sol.

(2) Inner solenoid is kept in uniform magnetic field of outer solenoid so force on all currect loops is zero Inner 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 additional mass M is added to its bob, the time period changes to TM. If the Young’s modulus of the material of the wire is Y then

1 is equal to : Y

(g = gravitational acceleration)

 TM  2  A (2)  T   1 Mg  

  T 2 A (1) 1       TM   Mg Sol.

(2) T  2

l g

Mg A

l

TM 59.

 2

l

  TM  2  A (4) 1   T   Mg  

...(1) ....(2)

l  l g

....(3)

From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its faces is : (1)

Sol.

Y

 TM  2  Mg (3)  T   1 A  

4MR2

(2)

3 3

MR2 32 2

(3)

MR2 16 2

(4)

4MR2 9 3

(4) From symmetry it is obvious that for maximum volume 3a  2R

Mass of cube =

60.

M

a3



4 R3 3 Mcubea2 I 6

2M 3

When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons Head Office : Pushpanjali Place, Boring Road Crossing, Opp. Alankar Place, Patna-1 Branch Office : Rukanpura Flyover, Near Sri Sai Yamaha, Bailey Road, Patna- 14 Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777


IITian’s TAPASYA

[ 14 ]

. . .I I Ti a n s c r e a t i n g I I T i a n s –4

–1

–28

is 2.5 x 10 ms . If the electron density in the wire is 8 x 10 close to :

Sol.

(1) 1.6 x 10 –5  m

(2) 1.6 x 10 –8  m

(1) I  neAVd, V

Il A



V neVdl

 IR 

–3

m , the resistivity of the material is

(3) 1.6 x 10 –7  m

(4) 1.6 x 10 –6  m .

 1.56  105



Jee Main Physics Solutions 2015

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