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(A) The moment of inertia of a solid sphere about an axis passing through its centre is $I_{cm} = \frac{2}{5} mR^2$. Using the parallel axis theorem,

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$265$

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(A) The moment of inertia of a solid sphere about an axis passing through its centre is $I_{cm} = \frac{2}{5} mR^2$. Using the parallel axis theorem, the moment of inertia about an axis at a distance $d$ from the centre is $I = I_{cm} + md^2$. $I = \frac{2}{5} mR^2 + md^2$. By definition, the radius of gyration $k$ is given by $I = mk^2$. Therefore, $mk^2 = \frac{2}{5} mR^2 + md^2$. $k^2 = \frac{2}{5} R^2 + d^2$. Given $R = 10 \ cm$ and $d = 15 \ cm$: $k^2 = \frac{2}{5} (10)^2 + (15)^2$. $k^2 = \frac{2}{5} (100) + 225 = 40 + 225 = 265$. Since $k = \sqrt{n}$, we have $k^2 = n$. Thus, $n = 265$.

$A$ solid sphere of radius $10 \ cm$ is rotating about an axis which is at a distance $15 \ cm$ from its centre. The radius of gyration about this axis is $\sqrt{n} \ cm$. The value of $n$ is

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