33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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64 NEET-AIPMT Chapterwise Topicwise Solutions Physics
Time taken by the man to complete one revolution is
T = 2π
= 2π
= 2π s
ωr
1
74. (d) : As no external torque is applied to the system,
the angular momentum of the system remains conserved.
\ L i = L f
According to given problem,
I
I t w i = (I t + I b )w
tωi
f or ω f =
...(i)
( It
+ Ib)
Initial energy, Ei = 1 2
Itω
i
...(ii)
2
1
2
Final energy, Ef = ( It + Ib)ω f
...(iii)
2
Substituting the value of w f from equation (i) in equation
(iii), we get
Final energy,
2 2 2
1 ⎛ Itωi
⎞ It
i
Ef = It + Ib
⎝
⎜ It
+ Ib
⎠
⎟ = 1 ω
( )
...(iv)
2
2 ( It
+ Ib)
Loss of energy, DE = E i – E f
2 2
1 2 1 It
ωi
= Itωi
−
(Using (ii) and (iv))
2 2 ( It
+ Ib) 2 ⎛ 2 ⎞ 2 ⎛ 2 2
ω
+ − ⎞
i It
ωi It IbIt It
= ⎜It
− ⎟ = ⎜
⎟
2 ⎝ ( It
+ Ib) ⎠ 2 ⎝ ( It
+ Ib)
⎠
1 IbIt
2
=
2 ( It
+ Ib) ω i
75. (d) : As no external torque is acting about the axis,
angular momentum of system remains conserved.
I 1 w 1 = I 2 w 2
2
I1ω1
Mr ω Mω
⇒ ω2
= = =
I
2
2 ( M + 2m) r ( M + 2m)
76. (c) : Applying conservation of angular momentum
I1
I 1 w = (I 1 + I 2 )w 1 or ω1
= ω
( I1+
I2)
77. (b) : According to conservation of angular
momentum, L = Iw = constant
Therefore, I 2 w 2 = I 1 w 1
2
I1ω1
Mr ω Mω
or ω2
= =
= .
I
2
2 ( M+
4m)
r M+
4m
78. (b) : When a child sits on a rotating disc, no external
torque is introduced. Hence the angular momentum of
the system is conserved. But the moment of inertia of the
system will increase and as a result, the angular speed
of the disc will decrease to maintain constant angular
momentum.
79. (b) : Required work done = – (K f – K i ) = 0 + K i = K i
= + = ⎛ 2
1 2 1 2 1
⎝ ⎜ 1 2 ⎞ v 1 2
Iω
mv mR
⎠
⎟ + mv
2 2 2 2 2
R 2
3 2 3
−2 2
= mv = × 100 × ( 20 × 10 ) = 3J
4 4
80. (d) : Using law of conservation of energy,
1 1
mg h = mvcm
2 + mr
2 ω
2
2 4
1
⇒
2 1 2
mg ssin30°= mvcm
+ mvcm
2 4
2 2
3 vcm
3 4
⇒ s =
⇒ s = ×
4 g sin 30°
4 1
10 ×
2
3
⇒ s = × 4 × 2 12
= = 24 . m
10 5
81. (b) : Translational kinetic energy, Kt = 1 2
mv
2
Rotational kinetic energy, K r = 1 I
2
ω
2
1 2 1 2 1 2 1
∴ Kt
+ Kr
= mv + Iω = + ⎛ 2 2
⎝ ⎜ 2 2⎞
mv mr ⎟ ⎛ 2 2 5 ⎠ ⎝ ⎜
7 2
∴ Kt
+ Kr
= mv
10
Kt
So, = 5 K + K 7
t
r
v
r
2
⎞
⎟
⎠
⎡ 2
⎤
I = mr
⎣
⎢
2 (for sphere)
5
⎦
⎥
82. (d) : Time taken by the body to reach the bottom
when it rolls down on an inclined plane without slipping
is given by
⎛ 2
k ⎞
2l
⎜1+
⎝ 2 ⎟
t =
R ⎠
g sinθ
Since g is constant and l, R and sinq are same for both
td
∴ =
ts
2
k
1+
d
2
R
=
2
k
1+
s
2
R
2
R
1+
2
2
R
R
1+
2 2
2
5R
⎛ R 2 ⎞
kd
= k s = R
⎝
⎜ ,
⎠
⎟
2 5
3 5 15
= × = ⇒ td
> ts
2 7 14
Hence, the sphere gets to the bottom first.
83. (a) : Acceleration of the solid sphere slipping down
the incline without rolling is
a slipping = gsinq...(i)
Acceleration of the solid sphere rolling down the incline
without slipping is
g g
arolling = sinθ
sinθ
=
k
1+
1+
2 = 5 2
7 g sinθ ...(ii)
2
R
5
⎛
For solid sphere, k 2
2 ⎞
⎜
=
⎝
2 ⎟
R 5 ⎠
Divide eqn. (ii) by eqn. (i), we get
arolling
= 5 aslipping
7