The moment of inertia of a hollow cylinder of mass $M$ and inner radius $R_{1}$ and outer radius $R_{2}$ about its central axis is

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

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

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

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

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Solution

The correct option is A $\frac{1}{2} M (R_{2}^{2} - R_{1}^{2})$
$\begin{matrix} b y t h e p r i n c i p l e o f e r p o s i t i o n, w e c a n i n e r t i a o f h o l l o w c y l i n d e r i s s u b s t r a c t i o n o f m o m e n t o f i n e r t i a o f f u l l s o l i d c y l i n d e r a n d i n n e r s o l i d c y l i n d e r I = I_{o u t} - I_{i n} N o w, f o r t h e m a s s o f t h e t w o p a r t s w e c a n u s e M_{o u t} = \frac{M}{π (R_{2}^{2} - R_{1}^{2})} \times π R_{2}^{2} M_{o u t} = \frac{M}{(R_{2}^{2} - R_{1}^{2})} \times R_{2}^{2} s i m i l a r l y m a s s o f i n n e r s o l i d c y l i n d e r M_{i n} = \frac{M}{(R_{2}^{2} - R_{1}^{2})} \times R_{1}^{2} N o w m o m e n t o f i n e r t i a i s g i v e n a s I = \frac{M_{o u t} R_{2}^{2}}{2} - \frac{M_{i n} R_{1}^{2}}{2} I = \frac{\frac{M}{(R_{2}^{2} - R_{1}^{2})} R_{2}^{4}}{2} - \frac{\frac{M}{(R_{2}^{2} - R_{1}^{2})} R_{1}^{4}}{2} I = \frac{\frac{M}{(R_{2}^{2} - R_{1}^{2})}}{2} (R_{2}^{4} - R_{1}^{4}) I = \frac{M (R_{2}^{2} - R_{1}^{2})}{2} \end{matrix}$
Hence,
option $(A)$ is correct answer.

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