1

INSTRUCTIONS TO CANDIDATES :

(i) There is 70 questions in this paper.

(ii) Question paper has two parts. In part A1 (Q. No. 1 to 60) each question has four alternatives, out ofwhich only one is correct. Choose the correct alternative and fill the appropriate bubble, as shown.

Q. No. 22 a c d

In part A2 (Q. No. 61 to 70) each question has four alternatives, out of which any number of alternative

(1, 2, 3 or 4) may be correct. You have to choose ALL correct alternatives and fill the appropriatebubbles, as shown.

Q. No. 64 a c

(iii) For Part A1, each correct answer carries 3 marks whereas 1 mark will be deducted for each wronganswer. In Part A2, you get 6 marks if all the correct alternatives are marked. No negative marks inthis part.

Answers & Solutionsforforforforfor

INDIAN ASSOCIATION OF PHYSICS TEACHERS

NATIONAL STANDARD EXAMINATION

IN PHYSICS (NSEP) 2019-20

Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005, Ph.: 011-47623456

DATE : 24/11/2019

Part : A-1

Only one out of four options is correct

1. A pendulum is made by using a thread of length300 cm and a small spherical bob of mass 100 g. It issuspended from a point S. The bob is pulled from itsposition of rest at O to the point A so that the linearamplitude is 25 cm. The angular amplitude in radianand the potential energy of the bob in joule at A arerespectively

(a) 0.10 and 0.10

(b) 0.083 and 0.01

(c) 0.251 and 2.94

(d) 0.083 and 0.24

Answer (b)

Sol. Angular displacement (in radian) Arc

radius

25 1rad

300 12 = 0.083 rad

h

300 cm

25 cm

A

S

O

Paper Code : 62

Time : 120 Minutes Max. Marks : 240

2

NSEP 2019-20

Potential energy

= mgh

= mg�(1 – cos)

= 0.1 × 10 × 3 (1 – cos 4.77°)

= 3 × 0.003463

= 0.01 joule

2. Consider the following physical expressions

(I) v2 ( : density, v : velocity)

(II)Y L

L (Y : Young's modulus, L : length)

(III)2

0

( : surface density of charge)

(IV) hrg (h : rise of a liquid in a capillary tube ofradius r)

The expressions having same dimensional formulaare

(a) I and II only (b) II and III only

(c) II, III and IV only (d) I, II and III only

Answer (d)

Sol. v2 [ML–3] [M0L2T–2] = [ML–1T–2]

1 21 2[ML T ][L]

[ML T ][L]

Y L

L

3. Two simple pendulums of lengths 1.44 m and 1.0 mstart swinging together in the same phase. The twowill be in phase again after a time of

(a) 6 second (b) 9 second

(c) 12 second (d) 25 second

Answer (c)

Sol. Time period of first pendulum 1

1.442

9.8 T

Time period of second pendulum 2

1.02

9.8 T

1

2

1.441.2

1 T

T

T1 = 1.2T

2

Particles will again comes in phase after n

oscillations of longer pendulum and (n + 1)oscillations of smaller pendulum.

nT1 = (n + 1)T

2

2 21.2 ( 1) n T n T

0.2n = 1

n = 5

Total time 1

1.44 5 2 3.14 1.25 2

9.8 3.14

t nT

= 12 s

4. A home aquarium partly filled with water slides downan inclined plane of inclination angle with respectto the horizontal. The surface of water in theaquarium

(a) remains horizontal

(b) remains parallel to the plane of the incline

(c) forms an angle with the horizon where 0 < < (d) forms an angle with horizon, where < < 90

Answer (b)

Sol.

As aquarium slides down with acceleration gsinthen water in aquarium experience pseudo forcemgsin in backward direction parallel to incline. Sonet force parallel to incline is zero.

Hence to make surface perpendicular to resultantforce (w.r.t. surface) mgcos, surface of liquidbecomes parallel to inclined plane.

5. A sound source of constant frequency travels with aconstant velocity past an observer. When it crossesthe observer the sound frequency sensed by theobserver changes from 449 Hz to 422 Hz. If thevelocity of sound is 340 m/s, the velocity of thesource of sound is

(a) 8.5 m/s (b) 10.5 m/s

(c) 12.5 m/s (d) 14.5 m/s

Answer (b)

Sol. Apparent frequency when source approaches

observer, 1

s

vn n

v v

Apparent frequency just after crossing the

observer 2

s

vn n

v v

1

2

s

s

v vn

n v v

449

422

s

s

v v

v v

449v – 449vs = 422v + 422v

s

27v = 871 vs

27 34010.5 m/s

871

s

v

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NSEP 2019-20

6. Plots of intensity (I) of radiation emitted by a blackbody versus wavelength () at three differenttemepratures T

1, T

2 and T

3 respectively are shown

in figure. Choose the correct statement

T1

T2

T3

I

(a) T1 > T

2 > T

3 necessarily

(b) T3 > T

2 > T

1 necessarily

(c) T2 = (T

1 + T

3)/2 necessarily

(d) T22 = T

1 T

3 necessarily

Answer (a)

Sol. IT

1 T2

T3

1m

2m

3m

According to Wien's displacement law,

mT = b

1

m

T

1m

< 2m

< 3m

Hence, T1 > T

2 > T

3

7. Consider a composite slab consisting of two differentmaterials having equal thickness and equal area ofcross-section. The thermal conductivities are K and2K respectively. The equivalent thermal conductivityof the composite slab is

(a)2

3

K(b) 2K

(c) 3K (d)4

3

K

Answer (d)

Sol. In steady state, heat current will be same in series,so

1 2eff

1 2

1 2

d dK

d d

K K

2

11

2 2

d d d

d d d

K K K

2 2

3

K

4

3 K

8. A large horizontal uniform disc can rotate freelyabout a rigid vertical axis passing through itscentre O. A man stands at rest at the edge of thedisc at a point A. The mass of the disc is 22 timesthe mass of the man. The man starts moving alongthe edge of the disc. When he reaches A, aftercompleting one rotation relative to the disc, the dischas turned through

(a) 30°

(b) 90°

(c) 60°

(d) 45°

Answer (a)

Sol. Mass of disc = 22M

Mass of man = M

From conservation of angular momentum,

0M m D DI I � �

2 2122

2m DMR MR (M is mass of man)

rel

+ D = –11

D

rel

= –12D

12

rel

D

36030

12 12

rel

D

So, disc turns through 30° in the opposite sense ofmotion of man.

9. Two factories are sounding their sirens at 400 Hzeach. A man walks from one factory towards theother at a speed of 2 m/s. The velocity of sound is320 m/s. The number of beats heard by the personin one second will be

(a) 6 (b) 5

(c) 4 (d) 2.5

Answer (b)

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NSEP 2019-20

Sol. Factory 1 f = 400 Hz Factory 2f = 400 Hz

Observer

2 m/s

Apparent frequency received by observer from factory1 is

1

320 2400

320

318400

320

af

Apparent frequency received by observer fromfactory 2 is

2

320 2400

320

322400

320

af

Therefore, frequency of beat heard by observer will be

1 2

400| | (322 318) 5 Hz

320b a af f f

10. The temperature of an isolated blackbody falls fromT

1 to T

2 in time t. Then, t = Cx where x is

(a)2 1

1 1

T T

(b) 2 2

2 1

1 1

T T

(c) 3 3

2 1

1 1

T T

(d) 4 4

2 1

1 1

T T

Answer (c)

Sol. According to Stefan law the energy of thermalradiation emitted per unit time by a black body ofsurface area A is given by

4dQAT

dt

4dTmc AT

dt

(m is mass, c is specific heat capacity)

2

1

4

0

T t

T

mc dTdt

A T

2

1

4

T

T

mc dTt

A T

2

1

3

3

T

T

mct T

A

3 3

2 1

1 1t C

T T

where C is a positive constant.

11. Tow charges – q and – q are placed at points (0, d)and (0, –d). A charge +q, free to move along X axis,will oscillate with a force proportional to

(a) 2 2

1

x d (b) 2

1

x

(c) 32 2 2( )

x

d x(d) 2 2

1

x d

Answer (c)

Sol. Let us take a point P on x-axis at a distance x fromorigin

–qA

B

P

y

Y

(0, – )d–q

Xx

(0, )d

O

E1[due to –q (0, d)]

2(along )

kqPA

y

����

2 2(along )

kqE PB

y

����

1 2| | | |

netE E E � � �

= 2 cosE [∵ sine component

cancel each other]

22

kq x

yy 2 2 1/2[ ( ) ]y x d ∵

2 2 3/2

2

( )

kqx

x d

force on a charge placed at point P will be

net| | | |F q E� �

2 2 3/2

2

( )

q kqx

x d

2

2 2 3/22

( )

xkq

x d

2 2 3/2( )

xF

x d

12. The average translational kinetic energy of oxygen(M = 32) molecules at a certain temperature is0.048 eV. The tranlational kinetic energy of nitrogen(M = 28) molecules at the same temeprature is(consider the two gases to be ideal)

(a) 0.0015 eV (b) 0.042 eV

(c) 0.048 eV (d) 0.768 eV

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NSEP 2019-20

Answer (c)

Sol. Average translational kinetic energy per molecule ofan ideal gas of molecular mass m is

21

2 rmsK mv

2

31

2

Bk T

mm

3

2 Bk T

Hence, average translational kinetic energy K isproportional to absolute temperature and independentof molecular mass of the gas.

Therefore, average translational kinetic energy of boththe gases are same at the same temperature.

K = 0.048 eV

13. A concave mirror has a radius of curvature R andforms the image of an object placed at a distance1.5 R from the pole of the mirror. An opaque disc ofdiameter half the aperture of the mirror is placed withthe pole at the centre. As a result

(a) The position of the image will be the same butits central half will disappear

(b) The position of the image will be the same butits outer half will disappear

(c) The complete image will be seen at the sameposition and it will be exactly identical with theinitial image

(d) The complete image will be seen at the sameposition but it will not be identical in all respectwith the initial image

Answer (d)

Sol. Since half part is covered with opaque disc, theimage will be less brighter but the size and positionof image will remain same.

14. A ray of white light is made incident on the refractingsurface of a prism such that after refraction at thissurface, the green component falls on the secondsurface at its critical angle. The colours present inthe emergent beam will be

(a) Violet, indigo and blue.

(b) Violet, indigo, blue, yellow, orange and red.

(c) Yellow, orange and red.

(d) All colours

Answer (c)

Sol.A

i

White

Green

Red

Violet

sinc = 1

Cred

> Cgreen

> Cviolet

(Θ red

< green

< violet

)

Dispersion takes place at first refracting place.Using Snell’s law,

1

sinsin

ir

So, (r1)red

> (r1)green

> (r1)violet

(Θ red

< green

< violet

)

Now, angle of incidence at second suface,

r2 = A – r

1

So, (r2)violet

> (r2)green

> (r2)red

If (r2)green

= Cgreen

, then (r2)red

< Cred

(r2)orange

< Corange

and (r2)yellow

< Cyellow

So, violet, indigo and blue will suffer total internalreflection and yellow, orange and red colours will bepresent in the emergent beam.

15. In a compound microscope, having tube-length30 cm, the power of the objective and the eye-pieceare 100D and 10D respectively. Then themagnification produced by the microscope when thefinal image is at the least distance of distinct vision(25 cm) will be

(a) 55

(b) 64

(c) 77

(d) 90

Answer (c)

Sol.

30 cm

u1

v1

u

f = 1 cm f = 10 cm

h

D

1 1 1

10

v u

v = –25 cm

1

160cm

7v

1

160cm

153u

1 1( / ) ( / )

/

hv u v u

D

h D

153 25

77 50

m 77

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NSEP 2019-20

16. Parallel rays are incident on a glass sphere ofdiameter 10 cm and having refractive index 1.5. Thesphere converges these rays at a certain point. Thedistance of this point from the centre of the spherewill be

(a) 2.5 cm (b) 5 cm

(c) 7.5 cm (d) 12.5 cm

Answer (c)

Sol.

P1 C

P2

5 cm

R = 5 cm = 3/2

Applying the refraction formula at first refractingsurface

2 1 2 1

1 2

31,

2

v u R

1

3/2 13/2( )

5

∵uv

1

1

3 115 cm

2 10 v

v

For second refracting surface

1 2

3, 1

2

u2 = +5 cm

2 1 2 1

v u R

1 3/21 32(5) 5

v

1 1 32.5 cm

10 10 v

v

Distance from centre of the sphere

= (2.5 + 5.0) = 7.5 cm

17. A jet of water from 15 cm diameter nozzle of a firehose can reach the maximum height of 25 m. Theforce exerted by the water jet on the hose is

(a) 4.24 kN (b) 17.32 kN

(c) 2.17 kN (d) 8.66 kN

Answer (d)

Sol.v 2

max 2 v

hg

25 × 19.6 = v2

v2 = 490

Force exerted by water jet on the hose is

F = Av2

= 3 415 1510 10 490 8.66 kN

4

18. In an electromagnetic wave the phase differencebetween electric vector and magnetic vector is

(a) Zero (b)2

(c) (d) 3

2

Answer (a)

Sol. In an electromagnetic wave, electric field vector andmagnetic field vector are perpendicular to each otherbut are in same phase. Hence phase differencebetween them is zero.

19. A spherical capacitor is formed by two concentricmetallic spherical shells. The capacitor is thencharged so that the outer shell carries a positivecharge and the inner shell carries an equal butnegative charge. Even if the capacitor is notconnected to any circuit, the charge will eventuallyleak away due to a small electrical conductivity ofthe material between the shells. What is thecharacter of the magnetic field produced by thisleakage current?

(a) Radially outwards from the inner shell to theouter shell.

(b) Radially inwards form the outer shell to the innershell.

(c) Circular field lines between the shells andperpendicular to the radial direction.

(d) No magnetic field will be produced

Answer (d)

Sol. As the current will be spherically diverging from thecentral sphere, we do not any consistent argumentfor the direction of magnetic field for any one of theelemental current. Because same argument will bevalid for other possible direction of magnetic field.

But if at all magnetic exist then it must have uniquedirection. So by argument of symmetry the magneticfield is zero.

20. If a cell of constant emf produces the same amountof heat during the same time in two independentresistors R

1 and R

2 when they are seperately

connected across the terminals of the cell, one afterthe other. The internal resistance of the cell is

(a) 1 2

2

R R(b) 1 2

2

R R∼

(c)2 2

1 2

2

R R(d)

1 2R R

Answer (d)

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NSEP 2019-20

Sol. For any cell

r

R

Heat developed in time (t) through a resistor R = H

2

2 EH I Rt Rt

R r

Given, H1 = H

2

2 2

1 22 2

1 2

E ER t R t

R r R r

2 2

2 1 1 2R r R R r R

2 2 2 2

2 2 1 1 1 22 2R r R r R R r R r R

2 2 2 2

1 2 1 2 1 1 2 2 1 22 2R R R r R R r R R R r R R r

2

1 2 1 2 1 2r R R R R R R

2

1 2r R R

1 2r R R

21. In the circuit shown beside the charge on eachcapacitor is

E2

E1

C1

C2

(a) (C1

+ C2)(E

1 – E

2)

(b) 1 2

1 2

1 2

CCE E

C C

(c) 1 2

1 2

1 2

CCE E

C C

(d) (C1

– C2)(E

1 + E

2)

Answer (c)

Sol. Let charge on each capacitor is 'q'

Applying KVL

E2

E1

C1

C2

–q +q

–q+q

2 1

2 1

0q q

E EC C

1 2

2 1

q qE E

C C

1 2

1 2

1 2

CCq E E

C C

22. A stationary hydrogen atom emits photoncorresponding to the first line (highest wavelength) ofLyman series. If R is the Rydberg constant and M isthe mass of the atom, the recoil velocity of the atomis

(a)4

Rh

M

(b)3Rh

M

(c)3

4

Rh

M

(d)Rh

M

Answer (c)

Sol. For first line of Lyman series

2 2

1 1 1 4

31 2R

R

Now for recoil velocity of atom

Photon

Mv

Pphoton

= Patom

hMv

3

4

h Rhv

M M

Hence 3

4

Rhv

M

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NSEP 2019-20

23. Heat is absorbed or evolved when current flows in aconductor having a temperature gradient. Thisphenomenon is known as

(a) Joule effect (b) Peltier effect

(c) Seeback effect (d) Thomson effect

Answer (d)

Sol. The phenomenon of absorption or evolution of heatalong the length of a conductor on passing currentthrough it when its two ends are kept at a differenttemperature is known as Thomson's effect.

24. Avalanche breakdown in a p-n junction primarilydepends on the phenomenon of

(a) Doping (b) Collision

(c) Recombination (d) Ionization

Answer (d)

Sol. The avalanche breakdown in a p-n junction diodeprimarily depends on the phenomenon called Impactionization.

25. A source emits photons of energy 5 eV which areincident on a metallic sphere of work function3.0 eV. The radius of the sphere is r = 8 × 10–3 m. Itis observed that after some time emission ofphotoelectrons from the metallic sphere is stopped.Charge on the sphere when the photoemission stopsis

(a) 1.77 × 10–16 C (b) 1.77 × 10–12 C

(c) 1.11 × 10–12 C (d) 1.11 × 10–10 C

Answer (b)

Sol. Energy of photons = h = 5 eV

Work function = = 3 eV

So, maximum kinetic energy of photoelectrons= (5 – 3) eV = 2 eV

Potential on the surface of sphere (when charge q is

attained) 0

1

4

q

r

Photoemission ceases when,

0

12

4

q

r

3

12

0 9

2 8 102 4 1.77 10 C

9 10q r

26. The dc component of current in the output of a half-wave rectifier with peak value l

0 is

(a) Zero (b) 0l

(c) 0

2

l

(d) 0

2l

Answer (b)

Sol. peak

0

1sin( )

2DCI I t dt

peak 0DC

I II

27. In an experiment on photoelectric effect, the slope of

straight line graph between the stopping potential andthe frequency of incident radiation gives

(a) Electron charge (e) (b) Planck constant (h)

(c)h

e(d) Work function (W)

Answer (c)

Sol. If incident radiation has frequency , and the workfunction of metal is and the stopping potential is V,then

h=+ eV

eV = h–

hV

e e

h

e is the slope of the mentioned graph.

28. According to Bohr's theory, the ionization energy ofH atom is 13.6 eV. The energy needed to remove anelectron from Helium ion (He+) is

(a) 13.6 eV (b) 16.8 eV

(c) 27.2 eV (d) 54.4 eV

Answer (d)

Sol.2

213.6 eV

n

ZE

n

Now, for He+ Z = 2, and for ground state n = 1

2 2

1 2

213.6 13.6 54.4 eV

1

ZE

n

So, ionization energy = 54.4 eV

29. The phenomenon inverse to photo electric effect is

(a) Compton effect

(b) Pair production

(c) Raman effect

(d) Production of X-rays in Coolidge tube

Answer (d)

Sol. In a photoelectric emission, electrons are emitted byincidence of photons, while in production ofX-rays in coolidge tube, photons are emitted usingincidence of high energy electrons. Thus productionof X-rays in coolidge tube is also termed as inversephotoelectric effect

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NSEP 2019-20

30. A stationary hydrogen atom emits a photon ofwavelength 1025 Å. Its angular momentum changesby

(a)h

(b)

2h

(c)2

h

(d)

3

2

h

Answer (a)

Sol. Let the electron transitions from n = 1 to n = n inthe atom

1

2

1 1109677 cm 1

n

8 2

1 1109677 1

1025 10 n

on solving, n 3

Now change in angular momentum 32 2

h h h

31. An observer stands on the platform at the front edgeof the first bogie of a stationary train. The train startsmoving with uniform acceleration and the first bogietakes 5 seconds to cross the observer. If all thebogies of the train are of equal length and the gapbetween them is negligible, the time taken by thetenth bogie to cross the observer is

(a) 1.07 s (b) 0.98 s

(c) 0.91 s (d) 0.81 s

Answer (d)

Sol.21 25

52 2

aa L L (i)

2 2

1 22 9 , 2 10v a L v a L

20 18aL aL at

20 18 25 18 25

202 2

L Lt

a a

5 10 3

= 0.81 s

32. The resistive force on an aeroplane flying in ahorizontal plane is given by F

f = kv2, where k is

constant and v is the speed of the aeroplane. Whenthe power output from the engine is P

0, the plane

flies at a speed v0 . If the power output of the engine

is doubled the aeroplane will fly at a speed of

(a) 1.12 v0

(b) 1.26 v0

(c) 1.41 v0

(d) 2.82 v0

Answer (b)

Sol. P = Fv

3

0 0P Kv

2P0 = Kv3

3

0

2v

v

1

3

0 02 1.26 v v v

33. A 3.0 cm thick layer of oil (density oil

= 800 kg/m3)floats on water (density

w = 1000 kg/m3) in a

transparent glass beaker. A solid cylinder is

observed floating vertically with 1

3of it in water and

1

3 in the oil. Oil is gently poured into the beaker

until the cylinder floats in oil only. The fraction of thesolid cylinder in oil now is

(a)3

5(b)

2

3

(c)3

4(d)

8

9

Answer (c)

Sol. 03 3

s w

L LL A A A

30 600 kg/m3

w

s

LsA

=

L'

0A

'

0

600 3

800 4

sL

L

34. A wedge of mass M rests on a horizontal frictionlesssurface. A block of mass m starts sliding down therough inclined surface of the wedge to its bottom.During the course of motion, the centre of mass ofthe block and the wedge system

(a) Does not move at all

(b) Moves horizontally with constant speed

(c) Moves horizontally with increasing speed

(d) Moves vertically with increasing speed

Answer (d)

Sol. The external force acting on the system of thewedge and the block is resultant of (M + m)gdownward and normal force on the wedge by thehorizontal surface, which is also vertical.

centre of mass moves vertically downward withincreasing speed (in general).

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NSEP 2019-20

35. A uniform circular disc rotating at a fixed angularvelocity about an axis normal to its plane andpassing through its centre has kinetic energy E. Ifthe same disc rotates with an angular velocity 2about a parallel axis passing through the edge, itskinetic energy will be

(a) 2E (b) 4E

(c) 10E (d) 12E

Answer (d)

Sol.2 2 2

21

2 2 4

MR MRE

(i)

2

2 23

2 2

MRl MR MR

2 2 2 21 3(2 ) 3

2 2E MR MR (ii)

E = 12 E

36. Light of wavelength 640 nm falls on a plane diffractiongrating with 12000 lines per inch. In the diffractionpattern on a screen kept at a distance of 12 cm fromthe grating, the distance of the second ordermaximum from the central maximum is

(a) 1.81 cm (b) 2.41 cm

(c) 3.62 cm (d) 7.25 cm

Answer (*)

Sol.

22.54 10

12,000d

dsin = 2

3sin

5

tanY

D

Y = 9 cm

(* None of the options is correct)

37. If the force acting on a body in inversely proportionalto its speed, the kinetic energy of the body varieswith time t as

(a) t0 (b) t1

(c) t2 (d) t –1

Answer (b)

Sol.mdv A

Fdt v

[where A = constant]

mvdv Adt

21

2mv At

38. As shown in the figure, a block of mass m is hungfrom the ceiling by the system of springs consistingof two layers. The force constant of each of thesprings is k. The frequency of the verticaloscillations of the block is

k k k

m

k k

(a)1

2 5

k

m (b)1 4

2 5

k

m

(c)1 5

2 6

k

m (d)1 6

2 5

k

m

Answer (d)

Sol. For top 3 springs, k1 = 3k

For bottom 2 spings, k2 = 2k

2

1 2

1 2

6 6

5 5eff

k k k kk

k k k

1 6

2 5

kf

m

39. Two simple harmonic motions are given by x1 = a sin

t + acost and x2 = asin

3

at cos t. The ratio

of the amplitudes of the first to the second and thephase difference between them respectively are

(a)3

and 2 12

(b)

3and

2 12

(c)2

and 123

(d)

3 and

2 6

Answer (a)

Sol. x1 = asint + acost 2 sin

4a t

2

2sin cot sin

63 3

a ax a t t t

Now, 1

2

2 3 3

2 22

3

A a

aA

Also, phase difference 1 2

4 6

12

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NSEP 2019-20

40. A particle is projected from the ground with a velocity

–1ˆ ˆ ms3 10v i j �

. The maximum height attained

and the range of the particle are respectively givenby (use g = 10 m/s2)

(a) 5 m and 6 m (b) 3 m and 10 m

(c) 6 m and 5 m (d) 3 m and 5 m

Answer (a)

Sol.uu

y

ux

H

gR

Given ˆ ˆ(3 10 ) m/su i j �

i.e. ux = 3 m/s

uy = 10 m/s

Also g = 10 m/s2

Maximum height

2

2

yu

Hg

2

105 m

2 10

Range 2 2 10 3

10

y

x

uu

g

= 6 m

41. A 20 cm long capillary tube stands vertically withlower end just in water. Water rises up to 5 cm. Ifthe entire system is now kept on a freely fallingplatform, the length of the water column in thecapillary tube will be

(a) 5 cm (b) 10 cm

(c) Zero (d) 20 cm

Answer (d)

Sol. When capillary is placed inside a system fallingfreely, then g

eff will be zero i.e. there is no force to

balance surface tension, hence water will rise incolumn up to full length.

Hence h = 20 cm

42. Position-time graph of a particle moving in a potentialfield is shown below. If the mass of the particle is1 kg its total energy is approximately

0

–15

–10

–5

0

5

10

15

2 4 6 8 10

t(s)

x(cm)

(a) 15.45 × 10–4 J (b) 30.78 × 10–4 J

(c) 7.71 × 10–4 J (d) 3.85 × 10–4 J

Answer (c)

Sol.

2 4 6 8 10 t(s)

–5

5

x(cm)

0

From the graph, equation of position in function of

time is given as 2 2(5 10 )sin

8x t

i.e. 2(5 10 )sin4

x t

Now total energy of the particle will be

2

22 2 21 1

1 5 102 2 4

E m A

E = 7.71 × 10–4 J

43. The log-log graph for a non-linear oscillator is shownbelow. Assuming the constants to have appropriatedimensions the relationship between time period (T)and the amplitude (A) can be expressed as

0 1 2

8

6

4

2

log (

)T

log ( )A

(a) T = 1000A2 (b) T = 4A1/2

(c) T = 4A2 + B (d) T = 8A3

Answer (a)

Sol. From the given graph, slope log 5 3

2log 1 0

T

A

Now from equation (y – y0) = m(x – x

0)

log(T) – 3 = 2[log(A)]

logT = log(A2) + log(1000)

logT = log(1000A2)

T = 1000A2

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NSEP 2019-20

44. In many situations the point source emitting a wavestarts moving, through the medium, with velocity Vgreater than the wave velocity in that medium. Insuch a case when source velocity (V) > wavevelocity (v), the wave front changes

(a) From spherical to plane

(b) From spherical to conical

(c) From plane to spherical

(d) From cylindrical to spherical

Answer (b)

Sol. For vsource

> v,

S

The spherical wavefronts i.e. originated at variouspositions of the source extends in three dimensionsand becomes conical in shape.

45. If the average mass of a smoke particle in an Indiankitchen is 3 × 10–17 kg, the rms speed of the smokeparticles at 27°C is approximately

(a) 2 cm/s (b) 2 m/s

(c) 2 km/s (d) None of these

Answer (a)

Sol. r.m.s speed of smoke particles is given by

23

rms 17

3 3 1.38 10 300

3 10

Bk Tv

m

6414 10 m/s

24.14 10 m/s

vrms

2 cm/s

46. Two wires, made of same material, one thick and theother thin are joined to form one composite wire. Thecomposite wire is subjected to the same tensionthroughout. A wave travels along the wire and passesthe point where the two wires are joined. The quantitywhich changes at the joint are

(a) Frequency only

(b) Propagation speed only

(c) Wavelength only

(d) Both propagation speed and wavelength

Answer (d)

Sol. Wave speed depend on the inertia and elasticproperty and frequency remains unchanged even ifthe wave passes from one medium to other medium.

So wavelength and speed will change.

47. The frequency of the third overtone of a closed endorgan pipe equals the frequency of the fifth harmonicof an open end organ pipe. Ignoring end correction,the ratio of their lengths l

open : l

close is

(a) 10 : 7 (b) 10 : 9

(c) 2 : 1 (d) 7 : 10

Answer (a)

Sol. L

For closed organ pipe

Here, closed

7

4

vf

L

And for open organ pipe fifth harmonic is open

5

2

vf

L

So open closed

5 7

2 4

v v

L L

open

closed

5 4

2 7

L

L

open

closed

10

7

L

L

48. A rectangular slab of glass of refractive index 1.5 isimmersed in water of refractive index 1.33 such thatthe top surface of the slab remains parallel to waterlevel. Light from a point source in air is incident onthe surface of water at an angle such that the lightreflected from the glass slab is plane polarised, theangle is

(a) 84.4° (b) 48.4°

(c) 56.3° (d) 53.1°

Answer (a)

Sol.

3

3

2

1 = 1

2

4

3

For air-water interface

4sin sin

3

sinsin

4

3

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4

3sin

2

24sin

3

But for water glass interface reflected light iscompletely polarized

3

2

3 3 9tan

2 4 8

2

2

9 sin

84

sin3

2 281 16sin sin

64 9

2 144sin

145

sin 84.4°

49. In a spectrometer the smallest main scale division is

1

3 of a degree. The total number of divisions on the

vernier scale attached to the main scale is 60 whichcoincide with the 59 divisions of the main circularscale. The least count of the spectrometer is

(a) 20

(b) 20

(c) 30

(d) 30

Answer (b)

Sol. 1 unit of main scale is 1

3

Now 1 unit of 59

vernier scale60

unit of main

scale

Least count of spectometer = 59 1

160 3

= 1 1

60 3

= 1 1

60 6060 3

= 20

50. White light is used to illuminate two slits in Young’sdouble slit experiment. Separation between the twoslits is b and the screen is at a distance D (>>b)from the plane of slits. The wavelength missing at apoint on the screen directly in front of one of theslits is

(a)2

2

3

b

D(b)

22b

D

(c)2

3

b

D(d)

2

2

b

D

Answer (c)

Sol.

P

2

b

Path difference 1

2 2 2b D D

2

2

11

2

bx D D

D

2

2

bx

D

if (2 1)2

x n then will be missing

2

(2 1)2 2

bn

D

2

(odd positive integer)

b

D

2

missing3

b

D

51. In an ink-jet printer, an ink droplet of mass m isgiven a negative charge q by a computer-controlledcharging unit. The charged droplet then enters theregion between two deflecting parallel plates of lengthL separated by distance d (see figure below) with aspeed v. All over this region there exists a uniformdownward electric field E (in the plane of paper).Neglecting the gravitational force on the droplet, themaximum charge that can be given to this droplet,so that is does not hit any of the plates, is

vE q

L

d

(a)2

2

mv L

Ed(b)

2

2

mv d

EL

(c)2 2

md

Ev L(d)

2 2mv L

Ed

Answer (b)

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NSEP 2019-20

Sol.L

tv

in x-direction

v

L

y

x2

d

In y-direction Eq

am

And for max. charge on drop

21

2 2

d Eq L

m v

2

2

EqLd

mv

2

2

mv dq

EL

52. A converging beam of light is pointing to P. Twoobservations are made with (i) a convex lens offocal length 20 cm and (ii) a concave lens of focallength 16 cm placed in the path of the convergentbeam at a distance 12 cm before the point P. It isobserved that

(a) In both cases the images are real

(b) In both cases the images are virtual

(c) For (i) the image is real and for (ii) the image isvirtual

(d) For (i) the image is virtual and for (ii) the imgeis real

Answer (a)

Sol. For concave lens if the virtual object would havebeen at a distance greater than 16 cm, then onlyit will make the virtual image. So concave lens willmake the real image.

And convex lens will make the real image for virtualobject.

53. Identify the rank in order from dimmest to thebrightest when all the identical bulbs are connectedin the circuit as shown below.

D

C

A B

+_

(a) A = B > C = D (b) A = B = C = D

(c) A > C > B > D (d) A = B < C = D

Answer (d)

Sol. 'C' and 'D' are subjected to same potential differenceV(let) so they are equally bright. 'A' and 'B' are both

subjected to same potential difference of 2

V

So A = B < C = D

54. The unit of magnetizing field is

(a) tesla

(b) newton

(c) ampere

(d) ampere turn/meter

Answer (d)

Sol. Magnetizing field 'H' is related with magnetic field Bas

∵0

BH

And 0.

2

n i

Bd

So 0

( )(length)

(Ampere turn)

B

(Ampere turn)

( )(length)

BH

B

unit of H is ampere turn/meter

55. A star undergoes a supernova explosion. Just afterthe explosion, the material left behind forms auniform sphere of radius 8000 km with a rotationperiod of 15 hours. This remaining materialeventually collapses into a neutron star of radius 4km with a period of rotation

(a) 14 s

(b) 3.8 h

(c) 0.021 s

(d) 0.0135 s

Answer (d)

Sol. Let the mass be M0.

2 2

0 1 1 0 2 2

2 2

5 5M R M R

2

2

2 2(8000) (4 4)

(15 hour) T

2 2

16 (15 hour)

(8000)T

T2 = 0.0135 (seconds)

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NSEP 2019-20

56. A number of identical absorbing plates are arrangedin between a source of light and a photo cell. Whenthere is no plate in between, the photo current ismaximum. Under the circumstances let us focus onthe two statements

(1) The photo current decreases with the increasein number of absorbing plates.

(2) The stopping potential increases with theincrease in number of absorbing plates.

(a) Statements (1) and (2) are both true and (1) isthe cause of (2)

(b) Statements (1) and (2) are both true but (1) and(2) are independent

(c) Statement (1) is true while (2) is not true and (1)and (2) are independent

(d) Statement (1) is true while (2) is not true and (2)is the effect of (1)

Answer (c)

Sol. Number of absorbing plate will reduce the intensityso photocurrent will decrease on increasing thenumber of plates.

But stopping potential will be depending on themaximum energy of photon.

57. In a nuclear reaction, two photons each of energy0.51 MeV are produced by electron-positronannihilation. The wavelength associated with eachphoton is

(a) 2.44 × 10–12 m

(b) 2.44 × 10–8 m

(c) 1.46 × 10–12 m

(d) 3.44 × 10–10 m

Answer (a)

Sol. hc

E

12 122.43 10 m 2.44 10 m

hc

E

58. In the circuit shown if an ideal ammeter is connectedbetween A and B then the direction of current andthe current reading would be (assume I

s remains

unchanged)

4R

A

2R

4R 2R

2R

4RA B

Is

Is

(a) B to A and 2

sI

(b) A to B and 4

sI

(c) B to A and 9

sI

(d) B to A and 3

sI

Answer (c)

Sol.

4R

2R4R 2R

2R

4R

Is

Is

I1

A

I Is–

1 2 –I I1 s

I1

I Is–

1

B

– 2RI1 – 6R(2I

1 – I

s) + 4R(I

s – I

1) = 0

4Is – 4I

1 = 2I

1 + 12I

1 – 6I

s

10Is = 18I

1

1

5

9s

I I

So the current in ammeter is 10

9 9

s s

s

I II I

from B to A

59. In the following figures the velocity-time graphs forthree particles 1, 2 and 3 are shown.

+4

+2

0

–2

Velo

city (

m/s

)

0 2 4 6 8 10

time (s)Particle 1

+4

+2

0

–2

Velo

city (

m/s

)

0 2 4 6 8 10

time (s)Particle 2

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NSEP 2019-20

+4

+2

0

–2

Velo

city (

m/s

)

0 2 4 6 8 10

time (s)Particle 3

The magnitude of average acceleration of the threeparticles, over 10 s, bear the relationship

(a) a1 > a

2 > a

3(b) a

2 > a

1 > a

3

(c) a3 > a

2 > a

1(d) a

1 = a

2 = a

3

Answer (d)

Sol. For particle (1), 1

2| |

10v

at

����

�

For particle (2), 2

2| |

10v

at

����

�

For particle (3), 3

2| |

10v

at

����

�

Clearly, 1 2 3

| | | | | |a a a � � �

60. The potential energy (U) of a particle moving in apotential field varies with its displacement (x) asshown below.

0

–5

–10

U (

J)

–15 –10 –5 0 5 10 15x (m)

The variation of force F(x) acting on the particle as afunction of x can be represented by

2

1

0

–1

–2

F (N)

x (m)–15 –10 –5 0 5 10 15

(Fig i)

2

1

0

–1

–2

F (N)

x (m)–15 –10 –5 0 5 10 15

(Fig ii)

2

1

0

–1

–2

F (N)

x (m)–15 –10 –5 0 5 10 15

(Fig iii)

2

1

0

–1

–2

F (N)

x (m)–15 –10 –5 0 5 10 15

(Fig iv)

(a) Fig (i)

(b) Fig (ii)

(c) Fig (iii)

(d) Fig (iv)

Answer (d)

Sol.du

Fdx

Force is of magnitude 1 N

And force is (+ve) for x = – 10 to x = 0

And force is (–ve) for x = 0 to x = 10.

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NSEP 2019-20

61. A pin of small length 'a' is placed along the axis ofa concave mirror of focal length f, at the distanceu(>f) from its pole. The length of its image is 'b'. Ifthe same object is placed perpendicular to its axisat the same distance u and the length of its imageis now 'c', then

(a)2

2( )

f

b au f

(b) c ab

(c) u f

c bf

(d)2 3

3( )

a f

bcu f

Answer (a, b, c, d)

Sol.1 1 1

v u f

uf

vu f

2

2( )

af

bu f

af

cu f

Substituting value of b, c in option we get all theoptions.

62. A thin rod of length 10 cm is placed along the axis ofa concave mirror of focal length 30 cm in such a waythat one end of the image coincides with one end ofthe object. The length of the image may be

(a) 7.5 cm (b) 12 cm

(c) 15 cm (d) 10 cm

Answer (a, c, d)

Sol. Case I : One of the ends is at centre and other inbetween centre and infinity

1 1 1

70 30V

1 1 1

30 70V

1 40, 52.5 cm

2100V

V

l = 7.5 cm

Case - II one of ends is at centre and other isbetween centre and pole

1 1 1, 75 cm

50 30V

V

l = 15 cm

63. The mass of an electron can be expressed as

(a) 0.512 MeV (b) 8.19 × 10–14 J/c2

(c) 9.1 × 10–31 kg (d) 0.00055 amu

where c is speed of light in vacuum

Answer (b, c, d)

Sol.1

amu 0.00055 amu1840

em

319.1 10 kg e

m

931 0.00055 0.512MeV e

m

31 16 2

14 2

9.1 10 9 10 J/c

8.19 10 J/c

e

m

64. Select the correct statement(s), out of the following,about diffraction at N parallel slits.

(a) There are (N – 1) minima between each pair ofprincipal maxima

(b) There are (N – 2) secondary maxima betweeneach pair of principal maxima

(c) Width of principal maximum is proportional to1/N

(d) The intensity at the principal maxima varies asN2

Answer (a, b, c, d)

Sol. (a) There use to be (N – 1) minima between eachpair of principal maxima.

(b) There would be (N – 2) secondary maximabetween each pair of principal maxima.

(c) For mth order and dsinm

= m

N cosm

m

Ad

width is proportional to 1

N

(d)2

2 2

0 2

sinN Nl l l

65. An electric dipole placed in a non-uniform electricfield may experience

(a) No net force, no torque

(b) A net force, but not torque

(c) No net force, but a torque

(d) A net force and a torque

Answer (a, b, c, d)

Sol. All the cases are possible.

Part : A-2

Any number of options (4), (3), (2) or (1) may be correct

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NSEP 2019-20

66. Two long parallel wires carry currents of equalmagnitude (I) but in opposite directions. These wiresare suspended from fixed rod PQ by four chords ofequal length L as shown. The mass per unit lengthof each wire is , the value of angle subtended bytwo chords OA and OB, assuming it to be small, is

Py O Q

C

B

x

A

z

L

I

I

(a)0

4

IgL

(b)0 1

IgL

(c) 0

4

gI

L(d) 0

gI

L

Answer (b)

Sol.

2

L L

T

Fm

Fg

2

0per unit length( )

2 2 sin2

m

IF

L

per unit length( )g

F g

cos2

T g

2

0sin

22 2 sin

2

lT

L

2

0

for small2

cos 12 2 2

sin2 2

l

L g

∵

0 1l

gL

67. A block of mass m = 10 kg is hanging over africtionless light fixed pulley by an inextensible lightrope. Initially the block is held at rest. The other endof the rope is now pulled by a constant force F in thevertically downward direction. The linear momentumof the block is seen to increase by 2 kg m/s in 1 s(in the first second). Therefore,

m

F

(a) The tension in the rope is F

(b) The tension in the rope is 3 N

(c) The work done by the tension on the block, infirst second, is = 19.8 J

(d) The work done against the force of gravity, in firstsecond, is = 9.8 J

Answer (a, d)

Sol. V = 0.2 m/s

a = 0.2 m/s2

20.20.1 m

2 0.2S

mg – F = ±ma

10F g a

Wg = (10 g × 0.1)

= 9.8 J

68. A ball of mass m1 travels horizontallly along the

x-axis in the positive direction with an initial speed ofv

0. It collides with another ball of mass m

2 that is

originally at rest. After the collision, the ball of massm

1 has velocity (v

1xi + v

1yj) and the ball of mass m

2

has velocity (v2x

i + v2y

j). Identify the correctrelationship(s)

(a) 0 = m1v

1x + m

2v

2x

(b) m1v

0 = m

1v

1y + m

2v

2y

(c) 0 = m1v

1y + m

2 v

2y

(d) m1v

0 = m

1v

1x + m

2v

2x

Answer (c, d)

Sol. Momentum of the system remains conserved inabsence of any external force

m1v

1y + m

2v

2y = 0

m1v

1x + m

2v

2x = m

1v

0

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NSEP 2019-20

69. In a real gas

(a) The force of attraction between the moleculesdepends upon intermolecular distance

(b) Internal energy depends only upon temperature

(c) Internal energy is a function of both temperatureand volume

(d) Internal energy is a function of both temperatureand pressure

Answer (a, c, d)

Sol. Variation of intermolecular forces as a function ofintermolecular distance

force

r

Internal energy of a real gas depends not just onlytemperature but also on volume and pressure aswell.

70. A particle of mass m is thrown vertically up withvelocity u. Air exerts an opposing force of a constantmagnitude F. The particle returns back to the pointof projection with velocity v after attaining maximumheight h, then

(a)2

2

uh

Fg

m

(b)2

2 –

vh

Fg

m

(c)

–

Fg

mv u

Fg

m

(d) v = u –

Fg

m

Fg

m

Answer (a, b, c)

Sol. During upward journey,

mg Fa

m

F

gm

mg

F

So, 2 20 2

Fu g h

m

2

2

uh

Fg

m

During downward journey

2

mg F Fa g

m m

mg

F

So, 2 20 2

Fv g h

m

Fg

mv u

Fg

m

�����