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
$\frac{I_1}{I_2}=\frac{{\left(\text{M}\right)}_1}{{\left(\text{M}\right)}_2}{\left(\frac{{\left(\text{R}\right)}_1}{{\left(\text{R}\right)}_2}\right)}^2=\frac{\lambda l_1}{\lambda l_2}{\left(\frac{{\left(\text{R}\right)}_1}{{\left(\text{R}\right)}_2}\right)}^2=\frac{2\pi \text{R}}{2\pi \text{nR}}{\left(\frac{\text{R}}{\text{nR}}\right)}^2$
$\lambda$ = linear density of wire = constant
$\Rightarrow \frac{I_1}{I_2}=\frac{1}{n^3}=\frac{1}{8}$ given
$\therefore n^3=8\Rightarrow n=2$