Question

is
.. From a uniform circular disc of radius R and mass
R
9M, a small disc of mass M and radius
3
removed concentrically. The moment of inertia of
the remaining disc about an axis perpendicular to
the plane of the disc and passing through its centre
is
1) 40/9 MR^2
2) MR^2
3) 4MR^2
4) 4/9 MR^2




​

Answer 1

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Answer 2

Explanation:

Given:

Mass of dics 9M 

Mass of removed part m=πR29M×π(R/3)3=M

Moment of inertia of Disc without hole about axis perpendicular to the plane passing through center is

I1=29MR2

Moment if inertia of hole  about axis passing through center of disc 

I2=2M(3R)2+M(32R)2

The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc

I=I1−I2

I=29MR2−⎣⎢⎢⎡2M(3R)2+M(32R)2⎦⎥⎥⎤=4MR2