Rotation 1 Question 12
15. From a solid sphere of mass $M$ and radius $R$, a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is
(a) $\frac{M R^{2}}{32 \sqrt{2} \pi}$
(b) $\frac{4 M R^{2}}{9 \sqrt{3} \pi}$
(c) $\frac{M R^{2}}{16 \sqrt{2} \pi}$
(d) $\frac{4 M R^{2}}{3 \sqrt{3} \pi}$
Show Answer
Answer:
Correct Answer: 15. (b)
Solution:
- Maximum possible volume of cube will occur when
$$ \begin{aligned} & \sqrt{3} a & =2 R \\ \therefore & a & =\frac{2}{\sqrt{3}} R \end{aligned} $$
Now, density of sphere, $\rho=\frac{M}{\frac{4}{3} \pi R^{3}}$
Mass of cube, $m=($ volume of cube $)(\rho)=\left(a^{3}\right)(\rho)$
$$ =\frac{2}{\sqrt{3}} R^{3} \frac{m}{\frac{4}{3} \pi R^{3}}=\frac{2}{\sqrt{3} \pi} M $$
Now, moment of inertia of the cube about the said axis is
