national standard examination in physics - rao … nsep_sol_2014.pdf · rao iit academy / nsep...

RAO IIT ACADEMY / NSEP Physics / Code : P 152 / Solutions

1

NATIONAL STANDARDEXAMINATION IN PHYSICS

2014 - 15

SOLUTIONS


2

NSEP SOLUTIONS (PHYSICS)CODE - P 152

ANSWER KEY & SOLUTIONS

1. (C) 2. (B) 3. (D) 4. (C) 5. (B)

6. (C) 7. (B) 8. (B) 9. (D) 10. (D)

11. (A) 12. (A) 13. (B) 14. (B) 15. (A)

16. (D) 17. (A) 18. (A) 19. (D) 20. (C)

21. (C) 22. (B) 23. (C) 24. (D) 25. (B)

26. (B) 27. (C) 28. (C) 29. (A) 30. (C)

31. (B) 32. (C) 33. (C) 34. (C)

35. (All Options are wrong) 36. (A) 37. (B) 38. (A)

39. (C) 40. (C) 41. (A) 42. (C) 43. (D)

44. (C) 45. (B) 46. (C) 47. (C) 48. (B)

49. (D) 50. (B) 51. (A) 52. (C) 53. (B)

54. (A) 55. (B) 56. (D) 57. (A) 58. (D)

59. (D) 60. (D) 61. (A, B) 62. (A, D)

63. (A, C, D) 64. (A, C, D) 65. (C) 66. (A, B, C, D)

67. (A, C) 68. (A, C) 69. (A, B, C, D) 70. (B, C)


3

SOLUTIONS

1. (C)

0

10log IdBI

0

120 10log II

12

0

10II

120 10II

2. (B)Dimension formula

0dEdt

is same is current density..

3. (D)r

m1 m2

x

22m r

21 222

Gm m m xr

2

T

21

2

2Gm xr T

2

1

2 r xTGm

1 2

1 2

m r m oxmm m

2

1

1 1 2

2 r m rTGm m m

3

1 2

2 rTG m m


4

4. (C)2 2

sN m R sin mg 2 2

eff smg mg m R sin 2 2

effg g R sin 2 2sin cos 1

2m Resin

Resin

2 2effg g R 1 cos

2 2 2k g e R cos

5. (B)22 3 4y x x

4 3dy xdx

0, 4at

3dydx

tan 3

1tan 3 71.56 72

6. (C)

1dydx

2 2V Vx Vy

Vx Vy

2 23 3 3 2V

7. (B)Thrust acting on the small element of width dx

2atmdF x g P r dx

2

0 0

22 atmgxF dF P x r

2

22 atm

gF P r

Thrust at bottom 2atmP g r

dx

x

10


5

10atmP g g

2

2 22atmgP g r Patm r

2

2 22gg g r g

1 1 22

r

10 102 2

r

5 10 15

8. (B)Equivalent length of a physical pendulum

Im cm

moment of inertia of a disc about it axis 2 21 2

12cmI M R R

So moment of inertia about outer radius22cmI I R M

R1

R2

2 2 21 2 2

12

I M R R M R

cm

IM

2 2 21 2 2

2

12

M R R MR

M R

2 221 21

2

2R R R

R

2 2 21 2 2

2

22

R R RR

2 21 2

2

32

R RR


6

9. (D)Final length of Aluminium

0 1 2 T

160 1 30 10 50A

160 1 10 10 50I

A I

t1 t2

A I

R

arcangleR

1 2

A I

R t R t

6 60 0

1 2

1 30 10 90 10 10HR t R t

4 41 15 10 1 5 10

4 41 15 10 1 5 101R R t

4 415 10 15 10R R t t 4 45 10 5 10R R t t

3 410 10 2 10 20R t t

3 410 2 10 20R

210 2.002

200.2cm

10. (D)

0 0 0 gg

2 0 0T g

12

Tn

mmg

mmg

T

T1

T

T2

g

2T mg g 1 0 0T mg g


7

12

Tn m

2 T

1 1

2 2

TT

0 0

0 0

2.2 gg

0

0

2.2

1

0

0

4.841

0 04.84 4.84

03.84 4.84

04.84 1.263.84

11. (A)

mg

sinmg

2

cosmr

2mr

cosmg

2

sinmr

N

F

2

tan 0.2rg

sin cos ' 2mmg F vr

2

sin cosmvmg Nr

2 2

sin cos cos cosmv mvmg mgr r

'2vv

2

tan 0.2vrg


8

2 2 2' 0.2 , 0.2 0.84

v v vg grg r r

sin cos 0.8 cos 0.8 cosg g g g

sin cos 0.8 cos 0.8cos

tan .8 0.8

0.2 1.08 0.8

1.08 0.6

0.6 0.551.08

12. (A)Force between two magnetic dipoles

H

CFr

Initial 41

C mgr

Final H2

C mgr

4241

rr

4

4

45

2560.41 0.4096625

13. (B)Electric field at a distance

02r E

r

2rForce at a distance x

02qgF Fx

work done to displaced from 2r r2

02

r

r

q dxx


9

2

0

22 r

q n

0

22q n

work done = change in K.E.

2

0

122 2q n m

2cosq n V

14. (B)

m

10 20 50

'm g

' 30 10m g mg

3 'm m

'm mass of scalem mass of body

m

20 50

'm g

mg

T

0 0 0T mg v g V g V g

0 0V g

12 ' 30T m g

0 0 12 30 'V g m g

0 0 0 012 10 10V m V

0 012 10

0 012 12 10

02 12

0 6


10

15. (A)Heat absorbed = Heat rejected.100 1 10 100 20sS

1 0.52sS

100 5 20 100 0.5 10

10 10.5 1 cos2 4

S s c

1 4.24

1 11.05KJ Kg k

16. (D)

2avs

RTVM

3rms

RTVM

8P

RTVM

rms avs pV V V

17. (A)1 36 10unit J

610 36 10unit J for one year

6 3 8036 10 365 36000 10100

m

456.2m JKgefficiency of engine

1

2

1 TMT

3001500

25

So the total coal req.5456.25 1140.6252


11

18. (A)Voltage (V) 640 10 50 62000 10 volt

3V 2 10 volt

3

30e e

V 2 10IR R

3

10e e

V 2 10IR 100 R 100

e

e

R 1003R

e e3R R 100 eR 50

3

302 10 2I

50 2

54 10 40 A 30 divisions 40 A

40 A 1.3 A1 divisions30 division division

19. (D)1.5 V

r

0.5 V 0.5 V

1.5 V r = 10m

4

C=1000 F

4r 10 Initial 0 4 4 6

1.5 1.5 100 150i10 100 10 10

0tq CE 1 e

RC

0q tE 1 ec RC

t0.5 1.5 1 e

RC

1 t1 e3 RC

t 1 2e 1RC 3 3

t 2In

RC 3

3t RC In 112


12

20. (C)++++

++

++

+ + + +

+ + + ++

++

++

+

+++

+2

2 c change gets non-uniformly distributed on the inner surface of the shell.

21. (C)

1 21000 2 1001 2g g

21

2

1001 1.0020011000

1 2 21 2 1 2t t

21

2 1

1 21 2

tt

6

6

1 30 101.002001

1 10 10tt

60.002001 10 100.15219.97999

t

22. (B)Ultrasonic is not connected

23. (C)

N

S

North

Bearth

Bmagnet

S

0 0

H magnet 33

M MB B4 r 4 20cm

0

magnet 3MB

4 40cm

3

magnet 40cm3

H

B 20 1B 40 8

Hmagnet

BB8


13

24. (D)2 2 2

03 4 5 5 2T

velocity of image3 4 5ˆ ˆ5 5 5

i j k

4

3

33 4ˆ ˆ5 5

i j k

25. (B)

0

1II2

0 22

II cos 302

0 2 2

3II cos 30 cos 602

0 0I 3 1 3I 0.0932 4 4 32

26. (B)

Glass = 1.5= 1.38

12 t 2n 12

550t 99.637 nm4 4 1.38

9t 99.637 10 m 60.099 10 m

t 0.10 m

27. (C)

Source

L L

MK

Capacitance doesnot correspond to spring constant “K”.


14

28. (C)2 2 2sourceRms

V 5 2 25 4 29

sourcerms

V 5.338

source rmsmax

V 2V 1.414 5.338

7.6

29. (A)

0tq CE 1 e

RC

3 6

2

RC 10 10 22 10RC 22 10 0.22

6 1.522 10 10 1 e

0.22

5 6.81822 10 1 e

5q 22 10 4

0q 22 10

0q q teC C RC

4

6

22 10 tV e22 10 0.22

t 330 ms

30. (C)Internal point

20

0 2

I r2 rB B rR

External point

01B 2 r I Br

B

31. (B)

V = 5 voltf = 50

I = 100 mA

Coil A Coil B

z sourceI VdIZ V


15

source

2 dI V100 I V

source2 V V

100

0

0

i sin wt

i w cos wt

MdE Mdt

32. (C)I the second case the two coils are in phase opposition and they have unequal number of turns.

33. (C)

stable

rod

N 3N 1

pressure 0N N te 12

ln2t

1 4 te1/ 2

ln 2t

t 1e4

t ln4

1/ 2 1/ 2ln4 ln4t t 2tln2

t 2 5730 11460

34. (C)Truth table

1 1 00 0 11 0 10 1 1

A B Y

It will at like NAND gate.

35. No Matching2 22 2ma ma a 2r aI 4 m m

12 4 2 2

ar2


16

22 2 2ma ma ma a aI 4 m12 4 4 2

222 2 2ma 1I ma ma ma 2

3

36. (A)Let e = coefficient of restitution. (conservation of energy)

21

12

mgh m

213.62

7.2u

7.2e collision h2

h1= 36cm

v

22

12

m mgh

22

21 7.2 0.362 10

eh e

Total distance 2 4 6 81 1 1 1 12h e h e h e h e h

2 4 6 80.36 1 .....e e e e

2

10.36 . .1

G Pe

2

11 0.361

me

21 0.36e

2 1 .036 0.64e

% loss in K.E in 1st impact

12

2 2

12

m u v

2

100mu

22

2

0.84100

uu

36%


17

37. (B)

2Tg

2 Fgm

02 2

2B rg

d

1 ,TA Bi

where A and B are constant as i increase, A - Bi decrease and hence T increase.

Correct option is (B)

38. (A)There is no difference in their frequencies.

39. (C)

For cut off ehcKE

124028000 eVnmeV

124028000

nm

0.044 nm

40. (C)Equation of the secondary cell must be less than that of the primary cell.

41. (A)Insulator

42. (C)

. . IR rP GL r R L

4.5 40.640 5

br R

20.4R

43. (D)Both are unimportant with respects to the poter trimeter.


18

44. (C)Current remains same througouts the wire.

45. (B)

2P has longer jockey distance.

46. (C)It is greater than the jockey will shown deflection. Si,ilarly if R is large current will not pass through theprimary circuit.

47. (C)For 1 cm deviation the voltage changes by 2.6 10V

3

3

6 1010

VIR

6 A

48. (B)Energy transfer maxm when frequency matches.

49. (D)Resonance occurs in maxm energy transfer and also in same phase.

50. (B)Same Frequency (F0).

51. (A)This pendulum having maxm amplitude has same Frequency and is in phase maxm 100%. Energy transferoccurs both are extreme positions

driverPE PE

12 2

1 2 21 1K A K A2 2

12 2 2 2

1 1 2 2 2M W A M W A

2 2M10 M8.16

100M M8.16

1.5018M M

1.5 M M

2K MW


19

52. (C)

LT 2g

1ln T ln 2 ln L ln g2

DT 1 DLT 2 L

2

2 LT 4g

1 LT T2 L

2

2

1 L 4 T2 g T

2gT

4

1DTT

53. (B)Damped oscillations

54. (A)rt

0A A e

0ln A ln A rt

55. (B)Resistive force Velocity from Stokes Law..

56. (D)2

3m

So, 230 203

V u = –30 cm 20

f

So, 1 1 120 30 f

50 1 ___ 1600 f

New position

So, 1 1 5 1

' 10 60 12V

' 10u

1 1 1 2 60' 12 10 120V

So virtual large at 60 cm from lens.


20

57. (A)Focal length of Concave lens = –24 cmFocal length of Convex lens = 12 cm

1 2

1 1 1 1 1 2 1 112 24 24 24eq

cmf f f

now, 30u cm ?v

24eqf cm

120 430

m

1 1 130 24v

1 1 124 30v

624 30

120v cm

58. (D)

17 cm

10 cm

2 24f cm f1 = 12

u1

30 cm 10 cm

1 1 1

1 1 1v u f

1

1 1 130 12v

1

1 1 30 12 18 112 30 30 12 30 12 20v

11

1 1 2020

vv

for Concave lens : 10u ?v

24f 1 1 1

10 24v

1 1 124 10v

17V So real & 57 cm from object.


21

59. (D) for lens 1.2Case 1 : in air

1 2

1 1 1

eqf f f

1 1rcleqf

1 [Constants]

1.2 1 [Constants]

1 .2eqf [Constants]

now magnitude in same but sign changes.

So, 1 .2rel

.8rel

1.2.8 M

1.5M

60. (D)When sunlight is coming for surface 1. [Refraction]

1

3 13 1 22 12V

1

3 1 __ 12 24V

from 2nd surface

1 2

1 1 11f R R

32

43

1 1 212 2 R

12R cm

1

4 24 3 3 3

3 2 12V V

4 1 13 24 72V


22

4 1 13 72 24V

4 43 72V

72 4 243 4

V cm a

for light going from water to airfor surface 1 :

3 42 3

1

3 4 / 32 12V

43

1

32

1

3 12 72V

Refraction from surface 23 3

2 2

1

1 112b V

1 1 172 24b

1 1 1 424 72 72b

18b cmSo option (D)

61. (A, B)Initial c.m. (1,1)

So, 0 2 0 212cm

m mxxm m m m

2 215

m mxm

m3(0,2)

m2m1

(2,0)3

25 2 2m m mx x

Same0 0 2 21

2cmm myy

m m m m

32y

So initial 3 32 2,

Final c.m. (2, 1)0 2 0 2 '2

5cmm mxx

m

10 2 2 ' ' 4m m mx x

0 2 0 2 ' 31 ' 25cmm myy y

m


23

final 34, 2

13 3,2 2

P

134, 2P

displacement 2 15 ˆ2

P P i

So magnitude 52 unit

Parallel to x axis.Option A, B

62. (A, D)Velocity of B when it reaches ‘A’.

2 2 2V u as 2 1 2 2 .16V

1 / secu m AB16 cm

1kg 2kgV = 1m/sec

22 / seca m g

1 .64 2 .36V

0.6 / secV m

now collision –0.6 / sec

beforecollision0 / sec

B

A

V mV m

So momentum conservation

'after collision

'A

B

VV

1 2 2 ' 'B A A BV V V V

0.6 2 ' ' ___ 1A BV V

' 'sap A B

app B A

V V VeV V V

' ' 0.6 ___ 2

2 ' ' 0.6 ___ 1A B

A B

V V

V V

3 ' 1.2AV

21.2' 0.4 / sec3AV m

So ' 0.2 / secBV m

So Rebounds


24

Distance travelled by(A) (B)

2 / seca g m 22 / seca m

0.4 / secAV m 0.2BV

So, 2

1 2AVda

2

2

0.22 2

d

.162 2

2 0.01d m

1 0.04d mSo separation = 0.04 + 0.01 = 0.05 M = 5 cmSo option A, D

63. (A, C, D)So, 1 2T T Mg ma mg maa g upward Option (A) torque about center of disc.

1 2MgT

23

2MgT

2 1T R T R

3

2 2mg mgR R

I mgR2

2mR mgR

2 option 2gR

option - DdL mgRdt

64. (A, C, D)

2oiBd

80.3 ˆ 2 10

2 3o

AB J J

0 80.3 ˆ 2 10

2 3CB J J


25

0 864 10

2 3BB i i

80 32 10

4 3DB i i

0 option AA CB B

8 82 10 2 2 10 5A BB B i J T

8 82 10 2 2 10A BB B i J T Option C

82 10A B C DB B B B i

So, (A,C,D)

65. (C)

1 2-3

-3 -31 2

V = 4m/sec V = 0.02m/sec25mm 2×10r = = 12.5×10 m r = = 10 m

2 2

1 1 2 2A V = A V

221 1 2 2r V n r V

2

1 1

2 2

r Vnr V

2 412.502

3125n Plants Option -CFor a plant

water volume 2 2 2 7200secA V t hour

2310 0.2 7200

6 2 210 2 10 72 10 6 3144 10 M

5 3452.16 10 M 4

34.52 10V M

3 31 10m tr 4 34.52 10 10V

14.52 10 tr


26

66. (A, B, C, D)The circuit is half wave rectifie.

Only Half wave will pass through.So, (B,C)Max output voltage will be V0

67. (A, C)For He Boiling poit of water 373k

Boiling poit of water N2=?

So, 1 1

2 2

T PT P

2

373 51T

2

373 74.65

T K

Vapour pressure of He & O2 will be different at same temperature so O2 will not give same result.

68. (A, C) for water is increased when tempwerature is decreased

T density soT density so so speed

So speed by right in medium will decrease.

&A C

69. (A, B, C, D)Example of fraunhofter diffraction as the screen is so farso all A,B,C,D.

70. (B, C)at max . 0 . maxy y K E P E

0 . max, 0y y K E PE

So B C

national standard examination in physics - rao … nsep_sol_2014.pdf · rao iit academy / nsep...

Documents