Question

The moment of inertia of a hollow sphere of mass $M$ having internal and external radii $R$ and $2R$ about an axis passing through its centre and perpendicular to its plane is :

• A. $\frac{3}{2}M{R}^2$
• B. $\frac{13}{32}M{R}^2$
• C. $\frac{31}{35}M{R}^2$
• D. $\frac{62}{35}M{R}^2$

Answer 1

The correct option is D $\frac{62}{35}M{R}^2$

Moment of Inertia of a spherical shell is

$I_{shell}=\frac{2}{3}m{r}^2$

The given hollow sphere has an inner radius $R$ and outer radius $2R$.

Volume $V=\frac{4}{3}\pi \left(\right(2R{)}^3-R^3)=\frac{4}{3}\pi \times 7R^3=\frac{28}{3}\pi R^3$

$\therefore \rho =\frac{M}{\frac{28}{3}\pi R^3}$

Consider a thin shell of radius $x$ and width $dx$ for the hollow sphere.

Mass of this shell $dm=\rho \times \left(4\pi x^{2}dx\right)$ where $4\pi x^{2}dx$ is the volume of the thin shell of width $dx$.

$\therefore dI=\frac{2}{3}dm{x}^2=\frac{2}{3}\rho \times \left(4\pi x^{2}dx\right)x^2$

$\therefore I={\int }_R^{2R}dI={\int }_R^{2R}\frac{2}{3}\times \rho \times \left(4\pi x^{2}dx\right)x^2=(\frac{2}{3}\times \frac{M}{\frac{28}{3}\pi R^3})\times \left(4\pi \right){\int }_R^{2R}{x}^{4}dx=\frac{3M}{7R^3}{\left(\frac{1}{5}{x}^5\right)|}_R^{2R}$

$\Rightarrow I=\frac{2\times 3M}{3\times 7R^3}\left(\right(2R{)}^5-R^5)=\frac{2\times 3M}{3\times 7R^3}\times \frac{31}{5}{R}^5$

$\Rightarrow I=\frac{62}{35}{R}^2$