Question

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

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

Answer 1

Answer: (A)

$2$

The moment of inertia $\left(I\right)$ of circular ring whose axis of rotation is passing through its center, $I_1=m_2{R}^2$
Also, $I_2=m_2(nR{)}^2$
Since, both rings have same density,
$\Rightarrow \frac{m_2}{2\pi \left(nR\right)\times A_2}=\frac{m_1}{2\pi R\times A_1}$
Where $A$ is cross-section of ring,
$A_1=A_2$ (given) $\therefore m_2=n{m}_1$
Given
$\frac{I_1}{I_2}=\frac{1}{8}=\frac{m_1{R}^2}{m_2(nR{)}^2}=\frac{m_1{R}^2}{n{m}_1(nR{)}^2}$
$\Rightarrow \frac{1}{8}=\frac{1}{n^3}$ or $n=2$