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

Answer 1

Answer: (B)

$MI$ of sphere $A$ about its diameter $I_{O^{\prime }}=\frac{2}{5}M{r}^2$
Now $MI$ 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)
$=4I_0=4\left[\frac{2}{5}M{r}^2+\frac{M{R}^2}{2}\right]$
$=\frac{2}{5}M\left[4r^2+5R^2\right]$