KEAM 2014 Physics Solutions for all Codes A1, A2, A3 & A4
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Transcript of KEAM 2014 Physics Solutions for all Codes A1, A2, A3 & A4
1
SCIENCE INSTITUTE KERAL ENGG. 2014 PHYSICS
EXPLANATION FOR A1 VERSION
1. 2
lg
T∝
⇒% error in g = % error in l + 2x % error in T
⇒1 × 2 + 2 × 3
⇒8%
3. Displacement of the dropped body +
Displacement of the thrown body = Total
height
2 21 1
2 2gt ut gt h+ − =
⇒ h
tu
= ⇒100
425
t s= =
Distance to the meeting point from top
= 21
2gt
= 4.9 × 42
= 78.4
5. In first 3 seconds, 21
2S ut at= + ⇒
12 = u × 3 + 1
2× a × 32
12 = 3u + 9
2a ⇒3 = u + 1.5 a -------(1)
In first 6 seconds,
42 = u × 6 + 216
2a× ×
7 = u + 3a ----------(2)
Solve (1) & (2) to get a
6. Impulse = change in P
= 2mv
Impulse
2mv =
7. L r P= ×
1 2 1
2 1 1
i j k
= −−
9. X component of velocity xdx
vdt
=
8xv t=
Y component of velocity 6yv t=
Resultant velocity 2 2x yv v v= +
10t=
11. Acceleration 2120 6010 /
6
fa m s
m
−= = =
Tension = ( )m g a+
= 2 (10 + 10) = 20 N
12. 27
10TE mv=
13. By work-energy theorem
2
21. .........(1)
2 4
vF x m v
= −
21.( ) .........(2)
2F x y mv+ =
(2) 4
(1) 3
x y
x
+⇒ =
4
3
xy x= −
2
( x = the initial thickness penetrated and y the
additional thickness penetrated)
15. Applying law of conservation of angular
momentum,
1 1 2 2 1 2( )I I I Iω ω ω+ = +
3 31 12 10 5 ( 10 )4I I− −× + × = +
31 0.5 10I −⇒ = ×
16. 0t
t
ω ωα −= = t
t
ω
= 02
t
πυ
=2
π
( 0υ = 1800 rpm = 1800
30,60
= t = 2 × 60 s)
17. Centripetal force = 2mv
r-------(1)
Angular momentum L = mvr
L
vmr
⇒ =
(1) ⇒2
.m L
CFr mr =
19. Energy required = 1
nMgR
n +
(here n = 1)
20. Total energy = − KE
21. Area of the big drop = Area of the droplets
2 2 21 24 4 4R r rπ π π= +
⇒ 2 21 2R r r= +
23. Pressure difference = Pressure due to 10 mm
of Hg = 1 cm of Hg
h gω ωρ = Pressure due to 10mm of Hg
Hg Hghhω
ω
ρρ
= = 1 13.6
13.61
× = cm
24. 21
2W kx=
21' (2 )
2W k x=
21' 3
2W W k x− = ×
3W=
3J=
25. 3
rms
RTV
M=
3
A
RT
mN=
26. 2
1
1T
Tη = − 2
1
1T
Tη⇒ = −
2
1
0.8T
T⇒ = 2 10.8T T⇒ =
In the second case
2
1
500.6
T
T
− =
T2 – 50 = 0.6 T1
0.8 T1 – 50 = 0.6 T1
0.2 T1 = 50
T1 = 250 K
T2 = 200 K
27. 2V
nC R=
n = no of degrees of freedom
28. Tree translational and two rotational degrees
of freedom
29. Comparing with standard equation of SHM,
we get 2ω =
2
Tπ
ω=
31. Here, given y = 0.707 A
2
Ay =
3
KE = 2 2 21( )
2m A yω −
Put ,2
Ay =
KE = 2
2 21( )
2 2
Am Aω −
=2
TE
33.
A1
N4N3N2N1
A2 A3
A1, A2 & A 3 = Antinodes
N1, N2, N3 & N4 = Nodes
34. In stretched string all harmonics are present.
35. Electric field near the sheet F
Eq
=
12
19
1.6 10
1.6 10E
−
−
×=×
= 107
But 02
Eσε
=
02 Eσ ε⇒ =
Total charge = Areaσ ×
02 E Areaε= ×
(The charge will be negative)
38. Centripetal force = Coulomb’s force
2
1 22
0
1.
4
q qmv
r rπε=
2 1 2
04
q qv
rmπε⇒ =
Period 2 r
Tv
π=
2 2
22
4 rT
v
π=
2 2
2 0
1 2
4 4r rmT
q q
π πε×=
39. Electric field F
Eq
=
1000
2= = 500
Potential difference = E × distance
= 500 × 2 × 10 – 2 = 10 V
40. Current through 4 Ω resistor near xy =
4
14
A=
Current through the 4 Ω resistor parallel to
4 Ω = 1 A
Total current in the circuit = 2A
Total potential difference = ( .)effI R×
= 2 × 8
= 16 V
41. PD across R = 2V
IR = 2
⇒ ( )10
2500
RR
=+
⇒ 125R = Ω
42. s pI I=
5
55
E Err R R
=+ +
r R⇒ =
43. Power loss = I2R P
IV
=
2P
RV =
⇒ Power loss 2
1
v∝
44. Charge in internal energy =I2Rt
= 12 × 50 × 2 × 60
= 6 kJ
4
46. 2I
TmB
π=
2 24I
TmB
π=
22
4I
mT
π=
2 5
2 5
4 9 109
16 1016
π
π
−
−
× ×=× ×
= 4 Am2
2 2 2
2
q B rKE
m= -----(1)
22
qBqB m
mυ π υ
π= ⇒ =
2 2 2 24
(1)2
m rKE
m
π υ⇒ =
2 2 2 22 m rπ υ=
49. (1 cos )W mB θ= − 0( 180 )θ =
52. V(max) = BANω
53. ps s
p p s
IN V
N V I= =
330 4
5524
p ps
s
N IN
I
× ×⇒ = = =
54. Average energy density of magnetic field =
= 20
1
2Eε
2
121 18.85 10
2 2− = × × ×
= 1
2× 4.43 × 10 – 12 = 2.2 × 10 – 12 J
57. tanµ ρ=
tan 60=
3=
1
sinCµ
= 1 1sin
3C −
⇒ =
60. sind nθ λ= here n = 2
2
sind
λθ = ⇒ 7
6
2 5 10sin
2.5 10θ
−
−
× ×=×
2
sin5
θ⇒ =
61. 1
2
h
mmKEλ λ= ⇒ ∝
62. Time for 50% decay = One half life
Time for 87.5% decay = Three half lives
Time interval between 50% decay and 87.5%
decay = Two half lives
63. 1 2
23 30R R A R A= = ∝
Area ∝ R2
2/3
(1) 125 25
(2) 27 9
Area
Area = =
64. KE hυ φ= −
' 2KE h υ φ= × −
2hυ φ= − ⇒more than twice the initial value
67. The out put .y A B C= +
68. Voltage gain = Current gain × resistance gain
= 49 × 10
2
= 245
71. LOS distance = 2 2Rh Rh+
8Rh=
5
KERAL ENGG. 2014 PHYSICS
EXPLANATION FOR A2 VERSION
55. 2
lg
T∝
⇒% error in g = % error in l + 2x % error in T
⇒1 × 2 + 2 × 3
⇒8%
57 Displacement of the dropped body +
Displacement of the thrown body = Total
height
2 21 1
2 2gt ut gt h+ − =
⇒ h
tu
= ⇒100
425
t s= =
Distance to the meeting point from top
= 21
2gt
= 4.9 × 42
= 78.4
59 In first 3 seconds, 21
2S ut at= + ⇒
12 = u × 3 + 1
2× a × 32
12 = 3u + 9
2a ⇒3 = u + 1.5 a -------(1)
In first 6 seconds,
42 = u × 6 + 216
2a× ×
7 = u + 3a ----------(2)
Solve (1) & (2) to get a
60. Impulse = change in P
= 2mv
Impulse
2mv =
61 L r P= ×
1 2 1
2 1 1
i j k
= −−
63. X component of velocity xdx
vdt
=
8xv t=
Y component of velocity 6yv t=
Resultant velocity 2 2x yv v v= +
10t=
65. Acceleration 2120 6010 /
6
fa m s
m
−= = =
Tension = ( )m g a+
= 2 (10 + 10) = 20 N
66. 27
10TE mv=
67 By work-energy theorem
2
21. .........(1)
2 4
vF x m v
= −
21.( ) .........(2)
2F x y mv+ =
(2) 4
(1) 3
x y
x
+⇒ =
4
3
xy x= −
( x = the initial thickness penetrated and y the
additional thickness penetrated)
69. Applying law of conservation of angular
momentum,
1 1 2 2 1 2( )I I I Iω ω ω+ = +
6
3 31 12 10 5 ( 10 )4I I− −× + × = +
31 0.5 10I −⇒ = ×
70. 0t
t
ω ωα −= = t
t
ω
= 02
t
πυ
=2
π
( 0υ = 1800 rpm = 1800
30,60
= t = 2 × 60 s)
71. Centripetal force = 2mv
r-------(1)
Angular momentum L = mvr
L
vmr
⇒ =
(1) ⇒2
.m L
CFr mr =
1. Energy required = 1
nMgR
n +
(here n = 1)
2. Total energy = − KE
3. Area of the big drop = Area of the droplets
2 2 21 24 4 4R r rπ π π= +
⇒ 2 21 2R r r= +
5 Pressure difference = Pressure due to 10 mm
of Hg = 1 cm of Hg
h gω ωρ = Pressure due to 10mm of Hg
Hg Hghhω
ω
ρρ
= = 1 13.6
13.61
× = cm
6 21
2W kx=
21' (2 )
2W k x=
21' 3
2W W k x− = ×
3W=
3J=
7 3
rms
RTV
M=
3
A
RT
mN=
8. 2
1
1T
Tη = − 2
1
1T
Tη⇒ = −
2
1
0.8T
T⇒ = 2 10.8T T⇒ =
In the second case
2
1
500.6
T
T
− =
T2 – 50 = 0.6 T1
0.8 T1 – 50 = 0.6 T1
0.2 T1 = 50
T1 = 250 K
T2 = 200 K
9 2V
nC R=
n = no of degrees of freedom
10. Tree translational and two rotational degrees
of freedom
11 Comparing with standard equation of SHM,
we get 2ω =
2
Tπ
ω=
13. Here, given y = 0.707 A
2
Ay =
KE = 2 2 21( )
2m A yω −
Put ,2
Ay =
KE = 2
2 21( )
2 2
Am Aω −
7
=2
TE
15
A1
N4N3N2N1
A2 A3
A1, A2 & A 3 = Antinodes
N1, N2, N3 & N4 = Nodes
16. In stretched string all harmonics are present.
17 Electric field near the sheet F
Eq
=
12
19
1.6 10
1.6 10E
−
−
×=×
= 107
But 02
Eσε
=
02 Eσ ε⇒ =
Total charge = Areaσ ×
02 E Areaε= ×
(The charge will be negative)
20. Centripetal force = Coulomb’s force
2
1 22
0
1.
4
q qmv
r rπε=
2 1 2
04
q qv
rmπε⇒ =
Period 2 r
Tv
π=
2 2
22
4 rT
v
π=
2 2
2 0
1 2
4 4r rmT
q q
π πε×=
21 Electric field F
Eq
=
1000
2= = 500
Potential difference = E × distance
= 500 × 2 × 10 – 2 = 10 V
22. Current through 4 Ω resistor near xy =
4
14
A=
Current through the 4 Ω resistor parallel to
4 Ω = 1 A
Total current in the circuit = 2A
Total potential difference = ( .)effI R×
= 2 × 8
= 16 V
23 PD across R = 2V
IR = 2
⇒ ( )10
2500
RR
=+
⇒ 125R = Ω
24 s pI I=
5
55
E Err R R
=+ +
r R⇒ =
25 Power loss = I2R P
IV
=
2P
RV =
⇒ Power loss 2
1
v∝
26. Charge in internal energy =I2Rt
= 12 × 50 × 2 × 60
= 6 kJ
28 2I
TmB
π=
2 24I
TmB
π=
22
4I
mT
π=
8
2 5
2 5
4 9 109
16 1016
π
π
−
−
× ×=× ×
= 4 Am2
2 2 2
2
q B rKE
m= -----(1)
22
qBqB m
mυ π υ
π= ⇒ =
2 2 2 24
(1)2
m rKE
m
π υ⇒ =
2 2 2 22 m rπ υ=
31. (1 cos )W mB θ= − 0( 180 )θ =
34 V(max) = BANω
35 ps s
p p s
IN V
N V I= =
330 4
5524
p ps
s
N IN
I
× ×⇒ = = =
36. Average energy density of magnetic field =
= 20
1
2Eε
2
121 18.85 10
2 2− = × × ×
= 1
2× 4.43 × 10 – 12 = 2.2 × 10 – 12 J
39 tanµ ρ=
tan 60=
3=
1
sinCµ
= 1 1sin
3C −
⇒ =
42 sind nθ λ= here n = 2
2
sind
λθ = ⇒ 7
6
2 5 10sin
2.5 10θ
−
−
× ×=×
2
sin5
θ⇒ =
43. 1
2
h
mmKEλ λ= ⇒ ∝
44 Time for 50% decay = One half life
Time for 87.5% decay = Three half lives
Time interval between 50% decay and 87.5%
decay = Two half lives
45 1 2
23 30R R A R A= = ∝
Area ∝ R2
2/3
(1) 125 25
(2) 27 9
Area
Area = =
46 KE hυ φ= −
' 2KE h υ φ= × −
2hυ φ= − ⇒more than twice the initial value
49. The out put .y A B C= +
50 Voltage gain = Current gain × resistance gain
= 49 × 10
2
= 245
53. LOS distance = 2 2Rh Rh+
8Rh=
9
KERAL ENGG. 2014 PHYSICS
EXPLANATION FOR A3 VERSION
32. 2
lg
T∝
⇒% error in g = % error in l + 2x % error in T
⇒1 × 2 + 2 × 3
⇒8%
34. Displacement of the dropped body +
Displacement of the thrown body = Total
height
2 21 1
2 2gt ut gt h+ − =
⇒ h
tu
= ⇒100
425
t s= =
Distance to the meeting point from top
= 21
2gt
= 4.9 × 42
= 78.4
36. In first 3 seconds, 21
2S ut at= + ⇒
12 = u × 3 + 1
2× a × 32
12 = 3u + 9
2a ⇒3 = u + 1.5 a -------(1)
In first 6 seconds,
42 = u × 6 + 216
2a× ×
7 = u + 3a ----------(2)
Solve (1) & (2) to get a
37 Impulse = change in P
= 2mv
Impulse
2mv =
38. L r P= ×
1 2 1
2 1 1
i j k
= −−
40. X component of velocity xdx
vdt
=
8xv t=
Y component of velocity 6yv t=
Resultant velocity 2 2x yv v v= +
10t=
42. Acceleration 2120 6010 /
6
fa m s
m
−= = =
Tension = ( )m g a+
= 2 (10 + 10) = 20 N
43. 27
10TE mv=
44. By work-energy theorem
2
21. .........(1)
2 4
vF x m v
= −
21.( ) .........(2)
2F x y mv+ =
(2) 4
(1) 3
x y
x
+⇒ =
4
3
xy x= −
( x = the initial thickness penetrated and y the
additional thickness penetrated)
46. Applying law of conservation of angular
momentum,
1 1 2 2 1 2( )I I I Iω ω ω+ = +
10
3 31 12 10 5 ( 10 )4I I− −× + × = +
31 0.5 10I −⇒ = ×
47. 0t
t
ω ωα −= = t
t
ω
= 02
t
πυ
=2
π
( 0υ = 1800 rpm = 1800
30,60
= t = 2 × 60 s)
48. Centripetal force = 2mv
r-------(1)
Angular momentum L = mvr
L
vmr
⇒ =
(1) ⇒2
.m L
CFr mr =
50 Energy required = 1
nMgR
n +
(here n = 1)
51. Total energy = − KE
52. Area of the big drop = Area of the droplets
2 2 21 24 4 4R r rπ π π= +
⇒ 2 21 2R r r= +
54. Pressure difference = Pressure due to 10 mm
of Hg = 1 cm of Hg
h gω ωρ = Pressure due to 10mm of Hg
Hg Hghhω
ω
ρρ
= = 1 13.6
13.61
× = cm
55. 21
2W kx=
21' (2 )
2W k x=
21' 3
2W W k x− = ×
3W=
3J=
56. 3
rms
RTV
M=
3
A
RT
mN=
57 2
1
1T
Tη = − 2
1
1T
Tη⇒ = −
2
1
0.8T
T⇒ = 2 10.8T T⇒ =
In the second case
2
1
500.6
T
T
− =
T2 – 50 = 0.6 T1
0.8 T1 – 50 = 0.6 T1
0.2 T1 = 50
T1 = 250 K
T2 = 200 K
58. 2V
nC R=
n = no of degrees of freedom
59. Tree translational and two rotational degrees
of freedom
60. Comparing with standard equation of SHM,
we get 2ω =
2
Tπ
ω=
62. Here, given y = 0.707 A
2
Ay =
KE = 2 2 21( )
2m A yω −
Put ,2
Ay =
KE = 2
2 21( )
2 2
Am Aω −
11
=2
TE
64
A1
N4N3N2N1
A2 A3
A1, A2 & A 3 = Antinodes
N1, N2, N3 & N4 = Nodes
65 In stretched string all harmonics are present.
66. Electric field near the sheet F
Eq
=
12
19
1.6 10
1.6 10E
−
−
×=×
= 107
But 02
Eσε
=
02 Eσ ε⇒ =
Total charge = Areaσ ×
02 E Areaε= ×
(The charge will be negative)
69. Centripetal force = Coulomb’s force
2
1 22
0
1.
4
q qmv
r rπε=
2 1 2
04
q qv
rmπε⇒ =
Period 2 r
Tv
π=
2 2
22
4 rT
v
π=
2 2
2 0
1 2
4 4r rmT
q q
π πε×=
70. Electric field F
Eq
=
1000
2= = 500
Potential difference = E × distance
= 500 × 2 × 10 – 2 = 10 V
71 Current through 4 Ω resistor near xy =
4
14
A=
Current through the 4 Ω resistor parallel to
4 Ω = 1 A
Total current in the circuit = 2A
Total potential difference = ( .)effI R×
= 2 × 8
= 16 V
72 PD across R = 2V
IR = 2
⇒ ( )10
2500
RR
=+
⇒ 125R = Ω
31. s pI I=
5
55
E Err R R
=+ +
r R⇒ =
32. Power loss = I2R P
IV
=
2P
RV =
⇒ Power loss 2
1
v∝
33. Charge in internal energy =I2Rt
= 12 × 50 × 2 × 60
= 6 kJ
35. 2I
TmB
π=
2 24I
TmB
π=
22
4I
mT
π=
12
2 5
2 5
4 9 109
16 1016
π
π
−
−
× ×=× ×
= 4 Am2
2 2 2
2
q B rKE
m= -----(1)
22
qBqB m
mυ π υ
π= ⇒ =
2 2 2 24
(1)2
m rKE
m
π υ⇒ =
2 2 2 22 m rπ υ=
38. (1 cos )W mB θ= − 0( 180 )θ =
41. V(max) = BANω
42. ps s
p p s
IN V
N V I= =
330 4
5524
p ps
s
N IN
I
× ×⇒ = = =
43 Average energy density of magnetic field =
= 20
1
2Eε
2
121 18.85 10
2 2− = × × ×
= 1
2× 4.43 × 10 – 12 = 2.2 × 10 – 12 J
46 tanµ ρ=
tan 60=
3=
1
sinCµ
= 1 1sin
3C −
⇒ =
49. sind nθ λ= here n = 2
2
sind
λθ = ⇒ 7
6
2 5 10sin
2.5 10θ
−
−
× ×=×
2
sin5
θ⇒ =
50 1
2
h
mmKEλ λ= ⇒ ∝
51 Time for 50% decay = One half life
Time for 87.5% decay = Three half lives
Time interval between 50% decay and 87.5%
decay = Two half lives
52. 1 2
23 30R R A R A= = ∝
Area ∝ R2
2/3
(1) 125 25
(2) 27 9
Area
Area = =
53. KE hυ φ= −
' 2KE h υ φ= × −
2hυ φ= − ⇒more than twice the initial value
56 The out put .y A B C= +
57 Voltage gain = Current gain × resistance gain
= 49 × 10
2
= 245
60 LOS distance = 2 2Rh Rh+
8Rh=
13
KERAL ENGG. 2014 PHYSICS
EXPLANATION FOR A4 VERSION
13. 2
lg
T∝
⇒% error in g = % error in l + 2x % error in T
⇒1 × 2 + 2 × 3
⇒8%
15 Displacement of the dropped body +
Displacement of the thrown body = Total
height
2 21 1
2 2gt ut gt h+ − =
⇒ h
tu
= ⇒100
425
t s= =
Distance to the meeting point from top
= 21
2gt
= 4.9 × 42
= 78.4
17. In first 3 seconds, 21
2S ut at= + ⇒
12 = u × 3 + 1
2× a × 32
12 = 3u + 9
2a ⇒3 = u + 1.5 a -------(1)
In first 6 seconds,
42 = u × 6 + 216
2a× ×
7 = u + 3a ----------(2)
Solve (1) & (2) to get a
18. Impulse = change in P
= 2mv
Impulse
2mv =
19 L r P= ×
1 2 1
2 1 1
i j k
= −−
21 X component of velocity xdx
vdt
=
8xv t=
Y component of velocity 6yv t=
Resultant velocity 2 2x yv v v= +
10t=
23 Acceleration 2120 6010 /
6
fa m s
m
−= = =
Tension = ( )m g a+
= 2 (10 + 10) = 20 N
24. 27
10TE mv=
25. By work-energy theorem
2
21. .........(1)
2 4
vF x m v
= −
21.( ) .........(2)
2F x y mv+ =
(2) 4
(1) 3
x y
x
+⇒ =
4
3
xy x= −
( x = the initial thickness penetrated and y the
additional thickness penetrated)
27 Applying law of conservation of angular
momentum,
1 1 2 2 1 2( )I I I Iω ω ω+ = +
3 31 12 10 5 ( 10 )4I I− −× + × = +
14
31 0.5 10I −⇒ = ×
28. 0t
t
ω ωα −= = t
t
ω
= 02
t
πυ
=2
π
( 0υ = 1800 rpm = 1800
30,60
= t = 2 × 60 s)
29. Centripetal force = 2mv
r-------(1)
Angular momentum L = mvr
L
vmr
⇒ =
(1) ⇒2
.m L
CFr mr =
31 Energy required = 1
nMgR
n +
(here n = 1)
32. Total energy = − KE
33. Area of the big drop = Area of the droplets
2 2 21 24 4 4R r rπ π π= +
⇒ 2 21 2R r r= +
35. Pressure difference = Pressure due to 10 mm
of Hg = 1 cm of Hg
h gω ωρ = Pressure due to 10mm of Hg
Hg Hghhω
ω
ρρ
= = 1 13.6
13.61
× = cm
36. 21
2W kx=
21' (2 )
2W k x=
21' 3
2W W k x− = ×
3W=
3J=
37 3
rms
RTV
M=
3
A
RT
mN=
38 2
1
1T
Tη = − 2
1
1T
Tη⇒ = −
2
1
0.8T
T⇒ = 2 10.8T T⇒ =
In the second case
2
1
500.6
T
T
− =
T2 – 50 = 0.6 T1
0.8 T1 – 50 = 0.6 T1
0.2 T1 = 50
T1 = 250 K
T2 = 200 K
39 2V
nC R=
n = no of degrees of freedom
40. Tree translational and two rotational degrees
of freedom
41. Comparing with standard equation of SHM,
we get 2ω =
2
Tπ
ω=
43. Here, given y = 0.707 A
2
Ay =
KE = 2 2 21( )
2m A yω −
Put ,2
Ay =
KE = 2
2 21( )
2 2
Am Aω −
15
=2
TE
45
.
A1
N4N3N2N1
A2 A3
A1, A2 & A 3 = Antinodes
N1, N2, N3 & N4 = Nodes
46. In stretched string all harmonics are present.
47. Electric field near the sheet F
Eq
=
12
19
1.6 10
1.6 10E
−
−
×=×
= 107
But 02
Eσε
=
02 Eσ ε⇒ =
Total charge = Areaσ ×
02 E Areaε= ×
(The charge will be negative)
50. Centripetal force = Coulomb’s force
2
1 22
0
1.
4
q qmv
r rπε=
2 1 2
04
q qv
rmπε⇒ =
Period 2 r
Tv
π=
2 2
22
4 rT
v
π=
2 2
2 0
1 2
4 4r rmT
q q
π πε×=
51. Electric field F
Eq
=
1000
2= = 500
Potential difference = E × distance
= 500 × 2 × 10 – 2 = 10 V
52 Current through 4 Ω resistor near xy =
4
14
A=
Current through the 4 Ω resistor parallel to
4 Ω = 1 A
Total current in the circuit = 2A
Total potential difference = ( .)effI R×
= 2 × 8
= 16 V
53. PD across R = 2V
IR = 2
⇒ ( )10
2500
RR
=+
⇒ 125R = Ω
54. s pI I=
5
55
E Err R R
=+ +
r R⇒ =
55. Power loss = I2R P
IV
=
2P
RV =
⇒ Power loss 2
1
v∝
56 Charge in internal energy =I2Rt
= 12 × 50 × 2 × 60
= 6 kJ
58. 2I
TmB
π=
2 24I
TmB
π=
22
4I
mT
π=
16
2 5
2 5
4 9 109
16 1016
π
π
−
−
× ×=× ×
= 4 Am2
2 2 2
2
q B rKE
m= -----(1)
22
qBqB m
mυ π υ
π= ⇒ =
2 2 2 24
(1)2
m rKE
m
π υ⇒ =
2 2 2 22 m rπ υ=
61 (1 cos )W mB θ= − 0( 180 )θ =
64. V(max) = BANω
65. ps s
p p s
IN V
N V I= =
330 4
5524
p ps
s
N IN
I
× ×⇒ = = =
66. Average energy density of magnetic field =
= 20
1
2Eε
2
121 18.85 10
2 2− = × × ×
= 1
2× 4.43 × 10 – 12 = 2.2 × 10 – 12 J
69 tanµ ρ=
tan 60=
3=
1
sinCµ
= 1 1sin
3C −
⇒ =
72 sind nθ λ= here n = 2
2
sind
λθ = ⇒ 7
6
2 5 10sin
2.5 10θ
−
−
× ×=×
2
sin5
θ⇒ =
1. 1
2
h
mmKEλ λ= ⇒ ∝
2 Time for 50% decay = One half life
Time for 87.5% decay = Three half lives
Time interval between 50% decay and 87.5%
decay = Two half lives
3. 1 2
23 30R R A R A= = ∝
Area ∝ R2
2/3
(1) 125 25
(2) 27 9
Area
Area = =
4 KE hυ φ= −
' 2KE h υ φ= × −
2hυ φ= − ⇒more than twice the initial value
7. The out put .y A B C= +
8 Voltage gain = Current gain × resistance gain
= 49 × 10
2
= 245
11 LOS distance = 2 2Rh Rh+
8Rh=