JEE MAIN - Physics (2018 (Offline) - No. 16)
From a uniform circular disc of radius R and mass 9M, a small disc
of radius R/3 is removed as shown in the figure. The moment of
inertia of the remaining disc about an axis perpendicular to the plane
of the disc and passing through centre of disc is :
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$${{37} \over 9}M{R^2}$$
$$4M{R^2}$$
$${{40} \over 9}M{R^2}$$
$$10M{R^2}$$
Explanation
Given that for a uniform circular disc the radius is R and mass 9M
$$\therefore$$ The area of uniform circular disc = $$\pi {R^2}$$
The radius of removed portion = $${R \over 3}$$
$$\therefore$$ The area of removed portion = $${{\pi {R^2}} \over 9}$$
So the mass of the removed portion = $${{9M} \over 9}$$ = M
If the moment of inertia of removed disc about the center of the circular disc of radius R is I1 then
I1 = $${1 \over 2}M{\left( {{R \over 3}} \right)^2} + M{\left( {{{2R} \over 3}} \right)^2}$$ = $${{M{R^2}} \over 2}$$
Moment of inertia of the whole disc, I2 = $${{9M{R^2}} \over 2}$$
Moment of inertia of the remaining disc, I = I2 - I1
$$\therefore$$ I = $${{9M{R^2}} \over 2} - {{M{R^2}} \over 2}$$ = $${4M{R^2}}$$
$$\therefore$$ The area of uniform circular disc = $$\pi {R^2}$$
The radius of removed portion = $${R \over 3}$$
$$\therefore$$ The area of removed portion = $${{\pi {R^2}} \over 9}$$
So the mass of the removed portion = $${{9M} \over 9}$$ = M
If the moment of inertia of removed disc about the center of the circular disc of radius R is I1 then
I1 = $${1 \over 2}M{\left( {{R \over 3}} \right)^2} + M{\left( {{{2R} \over 3}} \right)^2}$$ = $${{M{R^2}} \over 2}$$
Moment of inertia of the whole disc, I2 = $${{9M{R^2}} \over 2}$$
Moment of inertia of the remaining disc, I = I2 - I1
$$\therefore$$ I = $${{9M{R^2}} \over 2} - {{M{R^2}} \over 2}$$ = $${4M{R^2}}$$
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