33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Work, Energy and Power
Telegram @unacademyplusdiscounts
47
2
1 2 1 F
kt
52. (a) : Energy = Kx =
v = 2
2 2 K
m
∵
K A
K
EA
= 1
EB
=
dv
or 2E
Acceleration of the particle, a =
A
B
EB
2
dt
1 2k
1 k
53. (d) : Initial velocity u = 20 m/s; m = 1 kg
a = =
2 m t 2mt
Kinetic energy = maximum potential energy
1 2 1 2
W = mv − m() 0 ⇒ kt = 1 2 M × g × h
mv [Using (i)] ∴ P =
2 2
2
t
Initial kinetic energy = 1 mk mk
2
× 1× 20 = 200 J
Force on the particle, F = ma = = t
−12
/
2 2t
2
Mgh (max) = 200 J
Work done
58. (d) : Power, P =
\ h = 20 m.
Time taken
The height travelled by the body, h′ = 18 m
Here work done (= mgh) is same in both cases.
\ Loss of energy due to air friction
P1
t2
30 s 30 s 1
= mgh – mgh′
∴ = = = =
P2
t1
1 min 60 s 2
⇒ Energy lost = 200 J – 1 × 10 × 18 J = 20 J
59. (b) : P
54. (c) : Gain in potential energy = mgh
0 = Fv
= 2 × 10 × 10 = 200 J
∵ F = ma=
m dv
Gain in potential energy + work done against friction
dt
= work done = 300 J ∴ P = mv dv
0
\ Work done against friction = 300 – 200 = 100 J
dt
0dt = mvdv
55. (b) : Here, F t
v
^ ^
= ( 2ti+
3t 2 j)
N , m = 1 kg
Integrating both sides, we get ∫ Pdt 0 = m∫
vdv
Acceleration of the body,
F ( ti 2
0 0
2 ^
mv
2 + 3t j)
N
Pt
a = =
0 =
2
m 1 kg
Velocity of the body at time t,
Pt
v = ⎛ 12 /
v t
⎝ ⎜ 2 0 ⎞
^ ^ ^ ^ −
v = ∫ adt = ∫ ( 2ti+ 3t 2 j)
dt = t i+
t j m s 1
m ⎠
⎟ or ∝
60. (b) : Power,
F⋅ v = Fv cosθ
\ Power developed by the force at time t,
Just before hitting the earth q = 0°. Hence, the power
^ ^ ^ ^
P = F⋅ v = ( 2ti+ 3t 2 j) ⋅ ( t i+
t j ) W = (2t 3 + 3t 5 ) W exerted by the gravitational force is greatest at the instant
just before the body hits the earth.
56. (c) : Here, Volume of blood pumped by man’s heart,
V = 5 litres = 5 × 10 –3 m 3 ( 1 litre = 10 –3 m 3 61. (d) : Here,
)
Mass per unit length of water, m = 100 kg/m
Time in which this volume of blood pumps,
Velocity of water, v = 2 m/s
t = 1 min = 60 s
Power of the engine, P = mv
Pressure at which the blood pumps,
= (100 kg/m) (2 m/s)
P = 150 mm of Hg = 0.15 m of Hg
= 800 W
= (0.15 m)(13.6 × 10 3 kg/m 3 )(10 m/s 2 )
62. (d) : Power delivered in time T is
= 20.4 × 10 3 N/m 2
P = F·V = MaV
\ Power of the heart = PV or P = MV dV ⇒ PdT = MVdV
dT
t
3 2 −3 3
( 20. 4× 10 N/m )( 5×
10 m )
=
= 170 . W
⇒ PT = 1 MV
or P =
60 s
2 2 T
57. (c) : Constant power acting on the particle of mass
m is k watt.
63. (c) : Mass of water falling/second = 15 kg/s
h = 60 m, g = 10 m/s 2 , loss = 10% i.e., 90% is used.
or P = k; dW = k;
dW=
kdt
Power generated = 15 × 10 × 60 × 0.9 = 8100 W = 8.1 kW
dt
^ ^ ^ ^ ^ ^
W t
64. (b) : P = F⋅ v = ( 60 i+ 15 j −3k) ⋅( 2i − 4 j+
5k)
Integrating both sides,
∫ dW = ∫kdt
0 0
⇒ W = kt
Using work energy theorem,
…(i)
= 120 – 60 – 15 = 45 watts
work done
65. (b) : Power=
time taken = W t