Figure represents half of an annular disc of radii $R_{1}$ and $R_{2}$ . The mass of this half disc is $M$ . The moment of inertia of this disc about an axis passing through centre of full disc and perpendicular to the plane of disc is

\frac{M}{2} (R_{2}^{2} - R_{1}^{2})

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M (R_{2}^{2} + R_{1}^{2})

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\frac{M}{2} (R_{2}^{2} + R_{1}^{2})

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\frac{M}{2} \frac{(R_{2}^{4} + R_{1}^{4})}{(R_{2}^{2} - R_{1}^{2})}

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Solution

The correct option is D $\frac{M}{2} (R_{2}^{2} + R_{1}^{2})$
Taking an element of thickness dr at a distance r from the center ,
we have $d I = d m r^{2}$ where $d m = \frac{2 M}{π (R_{2}^{2} - R_{1}^{2})} π r d r$
Hence , $I = \int_{R_{1}}^{R_{2}} \frac{2 M}{R_{2}^{2} - R_{1}^{2}} r^{3} d r$
Hence , $I = \frac{M}{2} (R_{2}^{2} + R_{1}^{2})$

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