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(D) The moment of inertia of a solid sphere about an axis passing through its centre of mass is $I_{cm} = \frac{2}{5} MR^2$.According to the parallel

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$\frac{13}{20} MR^2$

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(D) The moment of inertia of a solid sphere about an axis passing through its centre of mass is $I_{cm} = \frac{2}{5} MR^2$.According to the parallel axis theorem,the moment of inertia $I$ about an axis at a distance $d$ from the centre is given by $I = I_{cm} + Md^2$.Here,$d = \frac{R}{2}$.Substituting the values,we get $I = \frac{2}{5} MR^2 + M\left(\frac{R}{2}\right)^2$.$I = \frac{2}{5} MR^2 + M\left(\frac{R^2}{4}\right)$.$I = \left(\frac{2}{5} + \frac{1}{4}\right) MR^2$.$I = \left(\frac{8 + 5}{20}\right) MR^2 = \frac{13}{20} MR^2$.

$A$ solid sphere of radius $R$ has mass $M$. The moment of inertia of the solid sphere about an axis at a distance $\frac{R}{2}$ from the centre is:

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