Question

Consider a sphere of mass $M$ and radius $R$ centered at origin. The density of material of the sphere is $\rho =A{r}^{\alpha }$, where $r$ is the radial distance, $\alpha$ and $A$ are constants. If the moment of inertia of the sphere about the axis passing through centre is $\frac{6}{7}M{R}^2$, then the value of $\alpha$ is

• A. 3
• B. 6
• C. 9
• D. -12

Answer 1

Answer: (D)

Given, density of sphere, $\rho =A{r}^{\alpha }$
(where, $r=$ radial distance and $A$ and $\alpha$ are constants)
Consider an elemental spherical shell of radius $r$ and thickness $dr$

Mass of elemental spherical shell,
$dm=$ Volume $\times$ Density
$dm=\left(4\pi r^2\right)dr\cdot A{r}^{\alpha }=4\pi a{r}^{2+\alpha }dr$
Mass of entire solid sphere,
$\begin{array}{rl}M& =4\pi A{\int }_0^R{r}^{2+\alpha }dr\\ M& =4\pi A{\left[\frac{R^{3+\alpha }}{3+\alpha }\right]}_{0^R}=\frac{4\pi A}{3+\alpha }\cdot R^{3+\alpha }\dots (\text{ ii })\end{array}$
Now, moment of inertia of elemental spherical shell is
$dI=\frac{2}{3}(dm)\cdot r^2=\frac{2}{3}\left(4\pi A{r}^{2+\alpha }dr\right)r^2$
Moment of inertia of entire solid sphere,
$<br/>I=\frac{2}{3}4\pi A{\left[\frac{r^{5+\alpha }}{5+\alpha }\right]}_0^R\Rightarrow I=\frac{2}{3}4\pi A\left[\frac{R^{5+\alpha }}{5+\alpha }\right]<br/>$
$<br/>\begin{array}{l}<br/>I={\int }_0^{R}dI=\frac{2}{3}4\pi A{\int }_0^R{r}^{A+\alpha }.\\ <br/>I=\frac{2}{3}4\pi A{\left[\frac{r^{5+\alpha }}{5+\alpha }\right]}_0^R\Rightarrow I=\frac{2}{3}\\ <br/>I=\frac{2}{3}\left(\frac{4\pi A}{3+\alpha }\cdot R^{3+\alpha }\right)\cdot \frac{R^2(3+\alpha )}{5+\alpha }\\ <br/>\text{ From Eq. (ii), we get }\\ <br/>I=\frac{2}{3}M{R}^2\cdot \left(\frac{3+\alpha }{5+\alpha }\right)\dots (\text{ iii })\\ <br/>\text{ It is given in the question. }<br/>\end{array}<br/>$
$<br/>\begin{array}{l}<br/>I=\frac{6}{7}M{R}^2(\text{ iv })\\ <br/>\text{ On comparing Eqs (iii) and (ov), we get }\\ <br/>\frac{2}{3}M{R}^2\cdot \left(\frac{3+\alpha }{5+\alpha }\right)=\frac{6}{7}M{R}^2\Rightarrow \frac{3+\alpha }{5+\alpha }=\frac{9}{7}\\ <br/>21+7\alpha =45+9\alpha \\ <br/>\Rightarrow \alpha =-12<br/>\end{array}<br/>$