Question

Two circular loops of radii $R$ and $nR$ are made of same wire. If their $M.I$ about their normal axis through centre are in the ratio $1:8$, the value of $n$ is:

• A. $6$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is C $2$

Given that,

Mass of one loop = $M$

Mass of n loop = $nM$

Radius of one loop = $R$

Radius of n loop = $nR$

Now, the moment of inertia is

$\frac{I_m}{I_{nm}}=\frac{1}{8}$

$\frac{m{R}^2}{nM\times n{R}^2}=\frac{1}{8}$

$\frac{1}{n^3}=\frac{1}{8}$

$n^3=8$

$n=2$

Hence, the value of $n$ is $2$