Question

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

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

Answer 1

Answer: (A)

$2$

Answer 2

Ratio of moment of inertia of the rings
$\frac{ I _{1}}{ I _{2}}=\left(\frac{ M _{1}}{ M _{2}}\right)\left(\frac{ R _{1}}{ R _{2}}\right)^{2}=\left(\frac{\lambda L _{1}}{\lambda L _{2}}\right)^{2}\left(\frac{ R _{1}}{ R _{2}}\right)^{2}$
$=\left(\frac{2 \pi R }{2 \pi n R }\right)\left(\frac{ R }{ nR }\right)^{2}$
$[\lambda=$ linear density of wire $=$ constant ]
$\Rightarrow \frac{ L _{1}}{ L _{2}}+\frac{1}{ n _{3}}+\frac{1}{8}$ (given)
$\therefore n ^{3}=8$
$\Rightarrow n =2$