Question

Four spheres, each of mass $M$ and radius $\omega$ are situated at the four corners of square of side $R$. The moment of inertia of the system about an axis perpendicular to the plane of square and passing through its centre will be

• A. $\frac{5}{2} M\left(4 r^{2}+5 R^{2}\right)$
• B. $\frac{2}{5} M\left(4 r^{2}+5 R^{2}\right)$
• C. $\frac{2}{5} M\left(4 r^{2}+5 r^{2}\right)$
• D. $\frac{5}{2} M\left(4 r^{2}+5 r^{2}\right)$

Answer 1

Answer: (B)

$M I$ of sphere $A$ about its diameter $I_{O^{\prime}}=\frac{2}{5} M r^{2}$
Now $M I$ of sphere $A$ about an axis perpendicular to the plane of square and passing through its centre will be

$I_{O}=I_{O^{\prime}}+M\left(\frac{R}{\sqrt{2}}\right)^{2}$
$=\frac{2}{5} M r^{2}+\frac{M R^{2}}{2}$
[by the theorem of parallel axis]
Moment of inertia of system (i.e. four sphere)
$=4 I_{0}=4\left[\frac{2}{5} M r^{2}+\frac{M R^{2}}{2}\right]$
$=\frac{2}{5} M\left[4 r^{2}+5 R^{2}\right]$