33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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Systems of Particles and Rotational Motion
61
The moment of inertia of complete disc about centre O
before removing the portion of the disc
IO = 1 2
MR
2
So, moment of inertia of the disc with removed portion is
I = I O – I′ O = 1 2 2
2 3MR
13MR
MR − =
2 32 32
47. (b) : Net moment of inertia of the system,
I = I 1 + I 2 + I 3 .
The moment of inertia of a shell about its diameter,
2 2
I1
= mr .
3
The moment of inertia of a shell about its tangent is given
by
2 2 2 2 5 2
I2 = I3 = I1
+ mr = mr + mr = mr
3
3
2
5 2 2 2 12mr
2
∴ I = 2 × mr + mr = = 4mr
3 3 3
48. (a) : According to the theorem of parallel axes,
I = I CM + Ma 2 .
As a is maximum for point B. Therefore I is maximum
about B.
49. (b) : According to the theorem of parallel axes, the
moment of inertia of the thin rod of mass M and length L
about an axis passing through one of the ends is
I = I CM + Md 2
where I CM is the moment of inertia of the given rod
about an axis passing through its centre of mass and
perpendicular to its length and d is the distance between
two parallel axes.
L
Here, ICM = I0,
d = 2
⎛ ⎞
∴ I = I + M
⎝
⎜
L 2
2
ML
0 ⎠
⎟ = I0
+
2 4
50. (d) : Moment of inertia for
the rod AB rotating about an axis
through the mid-point of AB
perpendicular to the plane of the
paper is Ml2
12 .
\ Moment of inertia about the axis through the centre
of the square and parallel to this axis,
⎛
I I Md M l 2
= + = +
l 2 ⎞ Ml 2
2
0 ⎝
⎜
⎠
⎟ =
12 4 3 ..
For all the four rods, I = 4 Ml
3
51. (d) :
Total mass = M, total length = L
Moment of inertia of OA about O = Moment of inertia
of OB about O
2 .
⇒ = × ⎛ ⎝ ⎜ ⎞
⎠
⎟ ⎛ 2 2
M.I M
⎝ ⎜
L⎞
. total 2
2 2⎠
⎟ ⋅ 1
3 = ML
12
52. (d) : Moment of inertia of a uniform circular disc
about an axis through its centre and perpendicular to its
2
plane is IC = 1 MR
2
\ Moment of inertia of a uniform circular disc about
an axis touching the disc at its diameter and normal to
the disc is I.
By the theorem of parallel axes,
I = I C + Mh 2 = 1 3
MR 2 + MR 2 = MR
2
2
2
53. (c) : Radius of gyration of disc about a tangential
5
axis in the plane of disc is R= K
2 1 , radius of gyration
of circular ring of same radius about a tangential axis in
the plane of circular ring is
3 K1
5
K2
= R ∴ = ⋅
2 K 6
2
54. (a) : Moment of inertia of a disc about its diameter
= 1 2
MR
4
Using theorem of parallel axes,
1 2 2 5 2
I = MR + MR = MR .
4
4
55. (c) : Moment of inertia of uniform circular disc
about diameter = I
According to theorem of perpendicular axes.
Moment of inertia of disc about axis
1
=
2
2I
= mr
2
Using theorem of parallel axes,
Moment of inertia of disc about the given axis
= 2I + mr 2 = 2I + 4I = 6I
56. (a) : Given: Angular acceleration, a = 3 rad/s 2
Initial angular velocity w i = 2 rad/s
Time t = 2 s
Using, θ= ω i t 1 + α
2
t 2
∴ θ = 2× 2+ 1 × 3× 4= 4+ 6=
10 radians
2
57. (b) : Given : Mass M = 2 kg, Radius R = 4 cm
Initial angular speed
π π
ω0 = 3rpm= 3× 2 rad/s = rad/s
60 10
We know that, w 2 = w 2 0 + 2aq
⇒ = ⎛ 2
0 ⎝ ⎜ π ⎞
10 ⎠
⎟ + 2 × α × 2π × 2π ⇒ α = − 1 2
rad/s
800
Moment of inertia of a solid cylinder,
× ⎛ 2
MR ⎝ ⎜ 4 ⎞
2 2
⎠
⎟
100 16
I = = =
2 2 4
10