Question

Two rings of radius $R$ and $nR$ made up of same material and same thickness have the ratio of moment of inertia about an axis passing through the centre as $1:8$ . The value of $n$ is

• A. $2$
• B. $2 \sqrt{2}$
• C. $4$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

The ratio of moment of inertia of the rings
$\frac{I_{1}}{I_{2}}=\frac{\left(\textit{M}\right)_{1}}{\left(\textit{M}\right)_{2}}\left(\frac{\left(\textit{R}\right)_{1}}{\left(\textit{R}\right)_{2}}\right)^{2}=\frac{\lambda l_{1}}{\lambda l_{2}}\left(\frac{\left(\textit{R}\right)_{1}}{\left(\textit{R}\right)_{2}}\right)^{2}=\frac{2 \pi \textit{R}}{2 \pi \textit{nR}}\left(\frac{\textit{R}}{\textit{nR}}\right)^{2}$
$\lambda $ = linear density of wire = constant
$\Rightarrow \frac{I_{1}}{I_{2}}=\frac{1}{n^{3}}=\frac{1}{8}$ given
$\therefore n^{3}=8\Rightarrow n=2$