A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of…

Question

A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is:

1. A. 7/8
2. B. 7/40
3. C. 7/57 (Correct answer)
4. D. 7/64

Correct Answer

Option C — 7/57

Detailed Solution & Explanation

The correct answer is $\dfrac{7}{57}$.

Key Points

• Let the big sphere ($2R$) have mass $M$: $I_{whole}=\dfrac{2}{5}M(2R)^2=\dfrac{8}{5}MR^2$.
• By uniform density, the cut-out sphere of radius $R$ has mass $M'=\dfrac{M}{8}$ (volume ratio $1:8$).
• Its centre is at distance $R$ from the $Y$-axis, so (parallel-axis) $I_{smaller}=\dfrac{2}{5}M'R^2+M'R^2=\dfrac{7}{5}\cdot\dfrac{M}{8}R^2=\dfrac{7}{40}MR^2$.
• $I_{rest}=I_{whole}-I_{smaller}=\left(\dfrac{8}{5}-\dfrac{7}{40}\right)MR^2$, so the ratio $=\dfrac{7/40}{8/5-7/40}=\dfrac{7}{64-7}=\dfrac{7}{57}$.