JAM Physics 2018 Solutions, Exams of Physics

This is jam 2018 Physics solutions prepared by fiziks

Typology: Exams

2018/2019

Uploaded on 01/08/2019

gul661998
gul661998 🇮🇳

5

(1)

2 documents

1 / 31

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
fiziks
InstituteforNET/JRF,GATE,IITJAM,M.Sc.Entrance,JEST,TIFRandGREinPhysics
H.No.40D,GroundFloor,JiaSarai,NearIIT,HauzKhas,NewDelhi110016
Phone:01126865455/+919871145498
Website:www.physicsbyfiziks.com|Email:[email protected]
1
IIT-JAM 2018 (Solutions)
Q1. – Q10. carry one mark each.
Q1. Let

33
,2
f
xy x y . The curve along which 20f
is
(a) 2
x
y (b) 2
x
y
(c) 6
x
y (d)
2
y
x
Ans.: (b)
Solution:

22
23333
22
220fxy xy
xy

 

2612
f
xy
20612 0fxy 2
x
y
Q2. A curve is given by

23
ˆ
ˆˆ
rt ti tj tk
. The unit vector of the tangent to the curve at
1t is
(a) ˆ
ˆˆ
3
ijk (b) ˆ
ˆˆ
6
ijk (c) ˆ
ˆˆ
22
3
ijk (d) ˆ
ˆˆ
23
14
ijk
Ans.: (d)
Solution: Let ˆ
n be a unit vector tangent to the curve at .t
22
ˆ
ˆˆ
/23
ˆ/14 5
dr dt i tj tk
ndr dt tt



at ˆ
ˆ23
ˆ
1, 14
ijk
tn

Q3. There are three planets in circular orbits around a star at distances ,4aa and 9a,
respectively. At time 0
tt, the star and the three planets are in a straight line. The period
of revolution of the closest planet is T. How long after 0
t will they again be in the same
straight line?
(a)
8T (b)
27T (c) 216T (d) 512T
Ans. : (c)
Solution: 3/2
1
Tka T ,

3/2
248Tka T ,

3/2
3927Tka T
Common time that all three star will meet again is 0123
216tTTT T
 , which is LCM
of all time period.
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a
pf1b
pf1c
pf1d
pf1e
pf1f

Partial preview of the text

Download JAM Physics 2018 Solutions and more Exams Physics in PDF only on Docsity!

Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics

H.No. 40 ‐D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐ 110016

Phone: 011 ‐26865455/+91‐ 9871145498 Website: www.physicsbyfiziks.com | Email: [email protected]

IIT-JAM 2018 (Solutions)

Q1. – Q10. carry one mark each.

Q1. Let (^)  

3 3 f x y ,  x  2 y. The curve along which

2  f  0 is

(a) x  2 y (b) x  2 y

(c) x  6 y (d) 2

y x

Ans.: (b)

Solution: (^)    

2 2 2 3 3 3 3 f (^) 2 x 2 y (^) 2 x 2 y 0 x y

2  f  6 x  12 y

2   f  0  6 x  12 y  0  x  2 y

Q2. A curve is given by (^)  

ˆ 2 ˆ^3 ˆ

r ttit jt k

. The unit vector of the tangent to the curve at

t  1 is

(a)

ijk (b)

ijk (c)

ijk (d)

ijk

Ans.: (d)

Solution: Let n ˆ be a unit vector tangent to the curve at t.

2 2

/ ˆ^2 ˆ 3 ˆ

dr dt i tj tk n dr dt (^) t t

 ^ at^

i j k t n

Q3. There are three planets in circular orbits around a star at distances a , 4 a and 9 a ,

respectively. At time tt 0 , the star and the three planets are in a straight line. The period

of revolution of the closest planet is T. How long after t 0 will they again be in the same

straight line?

(a) 8 T (b) 27 T (c) 216 T (d) 512 T

Ans. : (c)

Solution:

3/ 2 T 1  kaT ,  

3/ 2 T 2 (^)  k 4 a  8 T ,  

3/ 2 T 3 (^)  k 9 a  27 T

Common time that all three star will meet again is t 0 (^)  T 1 (^)  T 2 (^)  T 3 (^)  216 T , which is LCM

of all time period.

Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics

H.No. 40 ‐D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐ 110016

Phone: 011 ‐26865455/+91‐ 9871145498 Website: www.physicsbyfiziks.com | Email: [email protected]

Q4. A current I is flowing through the sides of an equilateral triangle of side a. The

magnitude of the magnetic field at the centroid of the triangle is

(a)

I

a

(b)

0 I

a

(c)

I

a

(d)

3 0 I

a

Ans.: (a)

Solution:

RSaaa and

RS

OS   a

For segment PQ

2sin 60 3 2 4 6

PQ QR RP

I I

B B B

a a

PQ

I

B B

a

Q5. Two vehicles A and B are approaching an observer O at rest with equal speed as shown

in the figure. Both vehicles have identical sirens blowing at a frequency f (^) s. The observer

hears these sirens at frequency f (^) A and f (^) B , respectively from the two vehicles. Which

one of the following is correct?

(a) f (^) Af (^) Bfs (b) f (^) AfBfs

(c) f (^) Af (^) Bfs (d) f (^) Af (^) Bfs

Ans.: (b)

Solution: Doppler shift

s A s s A

v f f v v

, (^) B s s s B

v f f v v

V AVBVs

f (^) Af (^) Bfs

R

P S Q

a (^) I O

A

B

O

A

B

O

Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics

H.No. 40 ‐D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐ 110016

Phone: 011 ‐26865455/+91‐ 9871145498 Website: www.physicsbyfiziks.com | Email: [email protected]

Q9. Which of the following arrangements of optical components can be used to distinguish

between an unpolarised light and a circularly polarised light?

(a) (b)

(c) (d)

Ans.: (c)

Solution: (i) In configuration (A), output will be linearly polarized for both

(ii) In configuration (B), output will be linearly polarized for both

(iii) In configuration (C), output will be linearly polarized of constant intensity if input is

unpolarised whereas it is linearly polarized with intensity varying from zero to maximum

if input is circularly polarized.

(iv) In configuration (D) output will be linearly polarized for both.

Q10. Which one of the following graphs shows the correct

variation of V 0 with Vi? Here, Vd is the voltage drop across

the diode and the OP-Amp is assumed to be ideal.

(a) (b)

(c) (d)

Ans. : (a)

Solution: During positive half cycle it behaves as voltage follower i.e. v 0 (^)  vi , during negative

half cycle v 0 (^)  0.

Light

plate analyser

Light

plate analyser

Light

polariser plate analyser

Light

polariser plate analyser

V i Vd V 0

RL

0 i

V

V 0

Vi

V 0

Vd

Vi

V 0

Vd

V 0

Vi 0

Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics

H.No. 40 ‐D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐ 110016

Phone: 011 ‐26865455/+91‐ 9871145498 Website: www.physicsbyfiziks.com | Email: [email protected]

Q11. – Q30. carry two marks each.

Q11. Which one of the figures correctly represents the TS diagram of a Carnot engine?

(a) (b)

(c) (d)

Ans. : (b)

Solution:

AB Isothermal expansion

BC Adiabatic expansion

CD Isothermal compression

DA Adiabatic compression

Q12. The plane of polarisation of a plane polarized light rotates by

0 60 after passing through a

wave plate. The pass-axis of the wave plate is at an angle  with respect to the plane of

polarization of the incident light. The wave plate and  are

(a)

0 , 60 4

(b)

0 , 2

(c)

0 , 2

(d)

0 , 4

Ans.: (b)

Solution: When plane polarized light is incident on the  / 4 plate, it converts it into circularly

polarized light, whereas  / 2 plate rotates is by angle 2  , where  is angle between

fast axis and polarization direction.

Given,

0 0 2   60   30.

T

S

T

S

T

S

T

S

P

V

A

B

D C

T

S

A B

D C

Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics

H.No. 40 ‐D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐ 110016

Phone: 011 ‐26865455/+91‐ 9871145498 Website: www.physicsbyfiziks.com | Email: [email protected]

Q14. The equation of state for one mole of a non-ideal gas is given by 1

B

PV A

V

, where

the coefficient A and B are temperature dependent. If the volume changes from V 1 to V 2

in an isothermal process, the work done by the gas is

(a) 1 2

AB

V V

(b)

2

1

ln

V

AB

V

(c) 2 1 1 2

ln

V

A AB

V V V

  ^   

(d) 2 1 1

ln

V V

A B

V

Ans. : (c)

Solution: 1

B

PV A

V

2

A AB

P

V V

2 2

1 1

2

V V

V V

A AB

W PdV dV dV V V

  (^)  

2

1 1 2

ln

V

A AB

V V V

Q15. An ideal gas consists of three dimensional polyatomic molecules. The temperature is

such that only one vibrational mode is excited. If R denotes the gas constant, then the

specific heat at constant volume of one mole of the gas at this temperature is

(a) 3 R (b)

R (c) 4 R (d)

R

Ans.: (c)

Solution: For a polyatomic gas

C (^) P  (^)  4  f (^)  R

CV  (^)  3  f (^)  R

As, f  1, CV  4 R

Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics

H.No. 40 ‐D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐ 110016

Phone: 011 ‐26865455/+91‐ 9871145498 Website: www.physicsbyfiziks.com | Email: [email protected]

Q16. A long solenoid is carrying a time dependent current such that the

magnetic field inside has the form (^)  

2 0 B tB t k ˆ

, where k ˆ is along

the axis of the solenoid. The displacement current at the point P on

a circle of radius r in a plane perpendicular to the axis

(a) is inversely proportional to r and radially outward

(b) is inversely proportional to r and tangential

(c) increases linearly with time and is tangential.

(d) is inversely proportional to

2 r and tangential

Ans.: (b)

Solution:

d B E dl dl dt

 ^  ^  

 (^) 

2

 E  2  r   2 B t 0   R

2 B 0 tR E r

d 0

E

J

t

2 0 0 1 d d

B R

J J

r r

Q17. Consider an ensemble of thermodynamic systems each of which is characterized by the

same number of particles, pressure and temperature. The thermodynamic function

describing the ensemble is

(a) Enthalpy (b) Helmholtz free energy

(c) Gibbs free energy (d) Entropy

Ans. : (c)

Solution: The variable

G  H  TS

dGdHTdSSdT

TdSVdPTdSSdT

dGVdPSdT

r (^) p

k^ ˆ

Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics

H.No. 40 ‐D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐ 110016

Phone: 011 ‐26865455/+91‐ 9871145498 Website: www.physicsbyfiziks.com | Email: [email protected]

Q20. Consider two waves y 1 (^)  a cos  tkz  and y 2 (^)  a cos    (^)  t  (^)  k   k (^)  z . The

group velocity of the superposed wave will be (    and  k  k )

(a)

 

k^ k

(b)

 

 

2 k k

(c) k

(d)

 

k^ k

Ans. : (c)

Solution: y 1 (^)  a cos  tkz , y 2 (^)  a cos^     t   k   kz 

1 2

2 cos cos 2 2 2

t kz k k y y y a t z

 ^ ^  ^   ^  ^   

v g k k

Q21. Consider a convex lens of focal length f^. A point object moves towards the lens along

its axis between 2 f and f. If the speed of the object is V 0 , then its image would move

with speed VI. Which of the following is correct?

(a) VIVo ; the image moves away from the lens.

(b) VI   V (^) o ; the image moves away from the lens.

(c) VIVo ; the image moves away from the lens.

(d) VIVo ; the image moves away from the lens.

Ans. : (c)

Solution:

2

0

I and

f V V f u v u f

For, uf , viV 0 and u decreases than v increases.

VIV 0 and image moves away from the lens.

Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics

H.No. 40 ‐D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐ 110016

Phone: 011 ‐26865455/+91‐ 9871145498 Website: www.physicsbyfiziks.com | Email: [email protected]

Q22. A disc of radius R 1 having uniform surface density has a concentric hole of radius

R 2 (^)  R 1. If its mass is M , the principal moments of inertia are

(a)

     

2 2 2 2 2 2 1 2 1 2 1 2 , , 2 4 4

M R  R M R  R M R  R

(b)

     

2 2 2 2 2 2 1 2 1 2 1 2 , , 2 4 4

M R  R M R  R M R  R

(c)

     

2 2 2 2 2 2 1 2 1 2 1 2 , , 2 4 8

M R  R M R  R M R  R

(d)

     

2 2 2 2 2 2 1 2 1 2 1 2 , , 2 4 8

M R  R M R  R M R  R

Ans. : (b)

Solution:

 

 

2 2

1 1

2 2 2 2 2 1 2 2 2 1

R R

zz ZZ R R

M M^ R^ R

I dm r r r dr I R R

 

I (^) xxI (^) yyIzz

By symmetry I (^) xxIyy. Therefore,

 

2 2 2 1

4

xx yy

M R R

I I

Q23. The function (^)  

x x f x x x

 ^ ^ 

^ ^ 

is expanded as a Fourier series of the form

a 0 (^) (^) n 1 an cos (^)  nx (^)  (^) n 1 bn sin nx

   

 (^)  . Which of the following is true?

(a) a 0 (^)  0, bn  0 (b) a 0 (^)  0, bn  0

(c) a 0 (^)  0, bn  0 (d) a 0 (^)  0, bn  0

Ans.: (b)

Solution:- (^)  

x x f x x x

 ^ ^ 

^ ^ 

 

0 (^0 )

a xdx xdx

 (^)   (^)     a 0  0

Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics

H.No. 40 ‐D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐ 110016

Phone: 011 ‐26865455/+91‐ 9871145498 Website: www.physicsbyfiziks.com | Email: [email protected]

Q25. The mean momentum p

of a nucleon in a nucleus of mass number A and atomic

number Z depends on A , Z as

(a)

1 p^ ^  A^3 (b)

1 p^ ^  Z^3 (c)

1 p A^3

 

(d) (^)  

2 p AZ^3

 

Ans. : (c)

Solution: The radius of a nucleus can be combined as 2

(greater than the wavelength of

electron)

The moment

h p

1/ 3

  R  R A 0 which implies

1/ 3

0

h p A R

  .

As,

1/ 3 p A

 

Q26. The Boolean expression (^)  AB (^)  AB (^)  AB can be simplified to

(a) AB (b) AB (c) AB (d) AB

Ans.: (c)

Solution: Y  (^)  AB (^)  AB (^)  AB  (^)  AB (^)  ABAB

 (^)  AB (^)  ABAB  ABABABAB

Q27. Consider the transformation to a new set of coordinates (^)   , (^) from rectangular Cartesian

coordinates (^)  x , y (^) , where   2 x  3 y and   3 x  2 y. In the (^)   , (^) coordinate system,

the area element dxdy is

(a)

d  d  (b)

d  d  (c) 5 d  d  (d)

d  d 

Ans.: (a)

Solution:-

 

 

J x y

J x y

x y

 

 

J x y J

J  

Since, area element in    system is,

dA  J d  d   d  d 

Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics

H.No. 40 ‐D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐ 110016

Phone: 011 ‐26865455/+91‐ 9871145498 Website: www.physicsbyfiziks.com | Email: [email protected]

Q28. A particle of mass m is in a one dimensional potential (^)  

, otherwise

x L V x

 ^ 

At some instant its wave function is given by (^)   1   2  

 x   x  i  x , where

 (^1)  x and  (^2)  x are the ground and the first excited states, respectively. Identify the

correct statement.

(a)

2 2

2

L

x E m L

(b)

2 2

2

L

x E m L

(c)

2 2

2

L

x E m L

(d)

2 2

2

L

x E m L

Ans.: (a)

Solution:

 

0 0 0

0

E

E E

E E

Where,

2 2

(^0 ) 2

E

mL

2 2 2 2

2 2

E

mL mL

1 1 2 1 1 2 2 1

i i

X   X    X    X    X 

L L

L

Q29. A raindrop falls under gravity and captures water molecules from atmosphere. Its mass

charges at the rate  m t (^)  , where  is a positive constant and m t  (^) is the instantaneous

mass. Assume that acceleration due to gravity is constant and water molecules are at rest

with respect to earth before capture. Which of the following statements is correct?

(a) The speed of the raindrop increases linearly with time

(b) The speed of the raindrop increases exponentially with time

(c) The speed of the raindrop approaches a constant value when  t  1

(d) The speed of the raindrop approaches a constant value when  t  1

Ans.: (c)

Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics

H.No. 40 ‐D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐ 110016

Phone: 011 ‐26865455/+91‐ 9871145498 Website: www.physicsbyfiziks.com | Email: [email protected]

SECTION - B

MULTIPLE SELECT QUESTIONS (MSQ)

Q. 31 – Q. 40 carry two marks each.

Q31. Let matrix

x M

. If det (^)  M (^)  0 , then

(a) M is symmetric (b) M is invertible

(c) one eigenvalue is 13 (d) Its eigenvectors are orthogonal

Ans.: (a), (c), (d)

Solution:- Since,

x M

If M  0  36  6 x  0  x  6

Hence,

x M

(a) Here, M M

  , so it is symmetric matrix

(b) Determinant (^)  M (^)   0 , so noninvertible matrix

(c) For eigenvalue-

M   I  0

(d) Eigen vectors for distinct eigen values for a symmetric matrix are orthogonal.

Q32. Let (^)  

6 2 f x  3 x  2 x  8. Which of the following statements is (are) true?

(a) The sum of all its roots is zero

(b) The product of its roots is

(c) The sum of all its roots is

(d) Complex roots are conjugates of each other.

Ans.: (a), (b), (d)

Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics

H.No. 40 ‐D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐ 110016

Phone: 011 ‐26865455/+91‐ 9871145498 Website: www.physicsbyfiziks.com | Email: [email protected]

Solution:- (^)  

6 2 f x  3 x  2 x  8

Now,

6 2 3 x  2 x  8  0

6 2 2 8 0 3 3

xx  

6 5 4 3 2  AxBxCxDxExFxG  0

xxxxxx  

Here, sum of roots 0

B

A

And product of roots

G

A

Since all coefficient are real, then complex roots are conjugate to each other.

Hence, options (a), (b) and (d) are correct.

Q33. Two projectiles of identical mass are projected from the ground with same

initial angle (^)   (^) with respect to earth surface and same initial velocity (^)  u

in the same plane. They collide at the highest point of their trajectories and

stick to each other. Which of the following statements is (are) correct?

(a) The momentum of the combined object immediately after collision is zero.

(b) Kinetic energy is conserved in the collision

(c) The combined object moves vertically downward.

(d) The combined object moves in a parabolic path.

Ans. : (a), (c)

Solution: At the highest point there is only Horizontal velocity. In horizontal direction there is

not any External force. So momentum in the horizontal direction is conserved.

(c) After collision whole system will full under gravitation.

Q34. Two beams of light in the visible range (^)  400 nm  700 nm interfere with each other at a

point. The optical path difference between them is 5000 nm. Which of the following

wavelengths will interfere constructively at the given point?

(a) 416.67 nm (b) 555.55 nm (c) 625 nm (d) 666.66 nm

Ans.: (a),(b) and (c)

u u

Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics

H.No. 40 ‐D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐ 110016

Phone: 011 ‐26865455/+91‐ 9871145498 Website: www.physicsbyfiziks.com | Email: [email protected]

Q36. Consider a convex lens of focal length f. The lens is cut along a diameter in two parts.

The two lens parts and an object are kept as shown in the figure. The images are formed

at following distances from the object:

(a) 2 f^ (b) 3 f^ (c) 4 f^ (d) 

Ans.: (b), (c) and (d)

Solution: For first lens

v u f

For second lens

v u f

(i) if u   , v  f , v   (ii) if u  2 f , v  2 f , v  2 f

(iii) if u  2 f , v  2 f No image (iv) if u  2 f , v  2 f , v  2 f

(v) if uf v ,  , v   (vi) uf v ,   ve , No image

Thus, V cannot be 2 f. The correct options are (b),(c) and (d)

Q37. Let the electric field in some region R be given by

2 2 Ee ^ y^ i ˆ^  ex ˆ j. From this we may

conclude that

(a) R has a non-uniform charge distribution

(b) R has no charge distribution

(c) R has a time dependent magnetic field.

(d) The energy flux in R is zero everywhere.

Ans.: (b), (c)

Solution:   E  0

 and   E  0

Thus R has no charge distribution and R has a time dependent magnetic field.

0 f

2 f x

Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics

H.No. 40 ‐D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐ 110016

Phone: 011 ‐26865455/+91‐ 9871145498 Website: www.physicsbyfiziks.com | Email: [email protected]

Q38. In presence of a magnetic field B ˆ j and an electric field (^)   E k (^)  ˆ, a particle moves

undeflected. Which of the following statements is (are) correct?

(a) The particle has positive charge, velocity ˆ

E

i B

(b) The particle has positive charge, velocity ˆ

E

i B

(c) The particle has negative charge, velocity ˆ

E

i B

(d) The particle has negative charge, velocity ˆ

E

i B

Ans.: (b), (d)

Solution: Fq ^ E  (^)  vB   0  

E

v B

For  ve charge: a   k ˆ

E

v x B

For  ve charge: ak ˆ

E

v x B

Q39. In a pn junction, dopant concentration on the p -side is higher than that on the n -side.

Which of the following statements is (are) correct, when the junction is unbiased?

(a) The width of the depletion layer is larger on the n -side.

(b) At thermal equilibrium the Fermi energy is higher on the p -side.

(c) In the depletion region, number of negative charges per unit area on the p -side is

equal to number of positive charges per unit area on the n -side

(d) The value of the built-in potential barrier depends on the dopant concentration.

Ans. : (a), (c) and (d)

Q40. Which of the combinations of crystal structure and their coordination number is (are)

correct?

(a) body centered cubic  8 (b) face centred cubic  6

(c) diamond  4 (d) hexagonal closed packed  12

Ans. : (a), (c), (d)