Question

A spherical portion A of radius \( R \) is removed from a solid sphere B of radius \( 2R \), both centered. The ratio of moments of inertia of remaining part to original is:

• A. \( 31:32 \)
• B. \( 7:8 \)
• C. \( 15:16 \)
• D. \( 4:7 \)

Hint:

Use volume ratios to scale mass, then apply \( I = \frac{2}{5}MR^2 \) appropriately when subtracting.

Answer 1

The Correct Option is A

Moment of inertia of solid sphere: \( I = \frac{2}{5}MR^2 \)
Let mass of large sphere be \( M \), then small sphere has mass \( \frac{M}{8} \) since volume ratio is \( \left( \frac{R}{2R} \right)^3 = \frac{1}{8} \)
Moment of removed part: \( I_A = \frac{2}{5} \cdot \frac{M}{8} \cdot R^2 = \frac{M R^2}{20} \)
Remaining moment: \( I_B - I_A = \frac{2}{5} M (2R)^2 - \frac{M R^2}{20} = \frac{8MR^2}{5} - \frac{MR^2}{20} = \frac{31MR^2}{20} \)
Original moment: \( \frac{32MR^2}{20} \Rightarrow \text{Ratio} = \frac{31}{32} \)