Rotational Motion Question 203
Question: 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 center and perpendicular to one of its faces is
Options:
A) $\frac{4MR^{2}}{9\sqrt{3\pi }}$
B) $\frac{4MR^{2}}{3\sqrt{3\pi }}$
C) $\frac{MR^{2}}{32\sqrt{2\pi }}$
D) $\frac{MR^{2}}{16\sqrt{2\pi }}$
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Answer:
Correct Answer: A
Solution:
[a] Here
$a=\frac{2}{\sqrt{3}}R$
Now, $\frac{M}{M’}=\frac{\frac{4}{3}\pi R^{3}}{a^{3}}$ $=\frac{\frac{4}{3}\pi R^{3}}{{{\left( \frac{2}{\sqrt{3}}R \right)}^{3}}}=\frac{\sqrt{3}}{2}\pi .$
$M’=\frac{2M}{\sqrt{3}\pi }$
Moment of inertia of the cube about the given axis,
$I=\frac{M’a^{2}}{6}$
$=\frac{\frac{2M}{\sqrt{3}\pi }\times {{\left( \frac{2}{\sqrt{3}}R \right)}^{2}}}{6}=\frac{4MR^{2}}{9\sqrt{3}\pi }$
