Question

A solid sphere of mass M, radius R has moment of inertia 'I' about its diameter. It is recast into a disc of thickness 't' whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains 'I'. Radius of the disc will be

• A. $\frac{4R}{11}$
• B. $\frac{3R}{4}$
• C. $\frac{2R}{\sqrt{15}}$
• D. $\frac{2R}{3}$

Hint:

Mass cancels out completely when recasting shapes! Setting up the raw coefficients directly ($\frac{2}{5}R^2 = \frac{3}{2}R'^2$) lets you cross-multiply and take the square root to isolate the new radius fraction in seconds.

Answer 1

The Correct Option is C

The moment of inertia of a solid sphere about its diameter is: $$I_{\text{sphere}} = \frac{2}{5}MR^2$$ Let the radius of the recast disc be $R'$. The moment of inertia of a flat disc about its central axis perpendicular to its plane is $\frac{1}{2}MR'^2$. Applying the parallel axis theorem ($I = I_{\text{cm}} + Md^2$) to find the moment of inertia about its edge axis ($d = R'$): $$I_{\text{disc}} = \frac{1}{2}MR'^2 + MR'^2 = \frac{3}{2}MR'^2$$ Since the moment of inertia remains unchanged ($I_{\text{sphere}} = I_{\text{disc}}$): $$\frac{2}{5}MR^2 = \frac{3}{2}MR'^2$$ $$\frac{2}{5}R^2 = \frac{3}{2}R'^2 \implies R'^2 = \frac{4}{15}R^2$$ $$R' = \frac{2R}{\sqrt{15}}$$
Final Answer:
The radius of the disc will be $\frac{2R}{\sqrt{15}}$, which corresponds to option (C).