NATIONAL STANDARD EXAMINATION IN PHYSICS - Rao … NSEP_Sol_2014.pdf · RAO IIT ACADEMY / NSEP...
Transcript of NATIONAL STANDARD EXAMINATION IN PHYSICS - Rao … NSEP_Sol_2014.pdf · RAO IIT ACADEMY / NSEP...
RAO IIT ACADEMY / NSEP Physics / Code : P 152 / Solutions
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NATIONAL STANDARDEXAMINATION IN PHYSICS
2014 - 15
SOLUTIONS
RAO IIT ACADEMY / NSEP Physics / Code : P 152 / Solutions
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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)
RAO IIT ACADEMY / NSEP Physics / Code : P 152 / Solutions
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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”.
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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
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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
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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%
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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.
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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
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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.
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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.
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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
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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
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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
RAO IIT ACADEMY / NSEP Physics / Code : P 152 / Solutions
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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
RAO IIT ACADEMY / NSEP Physics / Code : P 152 / Solutions
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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
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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