Two rings of radius $\mathrm{R}$ and $\mathrm{nR}$ made of same material have the ratio of moment of inertia…
Two rings of radius $\mathrm{R}$ and $\mathrm{nR}$ made of same material have the ratio of moment of inertia about an axis passing through centre is 1: 8 . The value of $\mathrm{n}$ is
2
$2 \sqrt{2}$
4
$1 / 2$
Solution
Ratio of moment of inertia of the rings
$
\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right)\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{2}
$
$\begin{aligned}=&\left(\frac{\lambda l_{1}}{\lambda l_{2}}\right)\left(\frac{R_{1}}{R_{2}}\right)^{2}=\left(\frac{2 \pi \mathrm{R}}{2 \pi \mathrm{n} \mathrm{R}}\right)\left(\frac{\mathrm{R}}{\mathrm{n} \mathrm{R}}\right)^{2} \\ &(\lambda=\text { linear density of wire }=\mathrm{constant}) \\ \Rightarrow \quad \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{1}{\mathrm{n}^{3}}=\frac{1}{8} \quad \text { (given) } \\ \therefore \quad \mathrm{n}^{3}=8 \Rightarrow \mathrm{n}=2 \end{aligned}$
