Question

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

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

Answer 1

Answer: (A)

$2$

Answer 2

$\frac{I_{1}}{I_{2}}=\left(\frac{M_{1}}{M_{2}}\right)\left(\frac{R_{1}}{R_{2}}\right)^{2}$
$=\left(\frac{\lambda l_{1}}{\lambda l_{2}}\right)\left(\frac{R_{1}}{R_{2}}\right)^{2}$
$=\left(\frac{\lambda l_{1}}{\lambda l_{2}}\right)\left(\frac{R_{1}}{R_{2}}\right)=\left(\frac{2 \pi R}{2 \pi n R}\right)\left(\frac{R}{n R}\right)^{2}$
$(\lambda=$ linear densityof wire $=$ constant $)$
$\Rightarrow \frac{I_{1}}{I_{2}}=\frac{I}{n_{3}}=\frac{1}{8}$ (given $)$
$\therefore n^{3}=8$
$\Rightarrow n=2$