Question

Density of a solid sphere is given by the expression $\rho ={\rho }_0(1+\frac{x}{R})$ where $x$ is the distance from the centre of the sphere and $0\le x\le R$, $R$ is the radius of the solid sphere. Moment of inertia of the solid sphere about an axis passing through its centre will be: (Given ${\rho }_0$ is a constant)

• A. $\frac{44}{45}\pi {\rho }_0{R}^5$
• B. $\frac{44}{45}\pi {\rho }_0{R}^2$
• C. $\frac{44}{35}\pi {\rho }_0{R}^2$
• D. $\frac{48}{38}\pi {\rho }_0{R}^5$

Answer 1

The correct option is A $\frac{44}{45}\pi {\rho }_0{R}^5$

Consider a small elemental hollow sphere of radius $x$ and thickness $dx$. Since solid sphere can be thought of as being made up of concentric hollow spheres, with limits of $x$ from $0\to R$


Moment of inertia of elemental hollow sphere about axis through centre,
$dI=\frac{2}{3}\left(dm\right)x^2...\left(i\right)$
where $dm$ is the mass of elemental hollow sphere given as
$dm=\rho dV...\left(ii\right)$
where $\rho \to$ density of the sphere
$dV\to$ Volume of elemental hollow sphere
$dV=\text{surface area of element}\times \text{thickness}\therefore dV=\left(4\pi x^2\right)dx...\left(iii\right)$
and $\rho ={\rho }_0(1+\frac{x}{R})$

From eq. $\left(ii\right)\&\left(iii\right)$,
$dm={\rho }_0(1+\frac{x}{R})\times \left(4\pi x^{2}dx\right)$

Substituting the value of $dm$ in Eq. $\left(i\right)$ we get,
$\underset{0}{\overset{I}{\int }}dI=\frac{2}{3}{\rho }_0\underset{0}{\overset{R}{\int }}(1+\frac{x}{R})\left(4\pi x^2\right)dx\times x^2$
$=\frac{8\pi }{3}{\rho }_0\underset{0}{\overset{R}{\int }}(1+\frac{x}{R})x^{4}dx$
$=\frac{8\pi }{3}{\rho }_0\underset{0}{\overset{R}{\int }}(x^4+\frac{x^5}{R})dx$
$=\frac{8\pi }{3}{\rho }_0{[\frac{x^5}{5}+\frac{x^6}{6R}]}_0^R$
$=\frac{8\pi {\rho }_0}{3}[\frac{R^5}{5}+\frac{R^5}{6}]$
$\therefore I=\frac{44\pi }{45}{\rho }_0{R}^5$