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 $d r$

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