Question

Three identical spherical shells, each of mass $m$ and radius $r$ are placed as shown in figure. Consider an axis $X{X}^{\prime }$, which is touching two shells and passing through diameter of third shell. Moment of inertia of the system consisting of these three spherical shells about $X{X}^{\prime }$ axis is :

• A. $\frac{11}{5}m{r}^2$
• B. $4m{r}^2$
• C. $\frac{16}{5}m{r}^2$
• D. $3m{r}^2$

Answer 1

The correct option is B $4m{r}^2$

Given,
Mass of a spherical shell $=m$
Radius of spherical shell $=r$


Using Parallel Axis Theorem,
Moment of Inertia (M.O.I) of shell $1$ about $X{X}^{\prime }$ $=\frac{2}{3}m{r}^2+m{r}^2=\frac{5}{3}m{r}^2...\left(1\right)$
Moment of Inertia of shell $2$ will be same as shell $1$ about $X{X}^{\prime }$ $=\frac{5}{3}m{r}^2....\left(2\right)$ [By symmetry]
Moment of Inertia of shell $3$ about $X{X}^{\prime }=\frac{2}{3}m{r}^2....\left(3\right)$
As, M.O.I about same axis are added together.
$\therefore$ On adding equation $1,2$ and $3$,
$I_{system}=\frac{5}{3}m{r}^2+\frac{5}{3}m{r}^2+\frac{2}{3}m{r}^2$
$\Rightarrow I_{system}=\frac{12}{3}m{r}^2$
$\Rightarrow I_{system}=4m{r}^2$
Moment of Inertia of the system about $X{X}^{\prime }$ is $4m{r}^2$