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Q.
Two rings of radii $R$ and $nR$ made up of same material have the ratio of moment of inertia about an axis passing through centre is 1 : 8. The value of $n$ is
JIPMERJIPMER 2012System of Particles and Rotational Motion
Solution:
The moment of inertia of circular ring whose axis of rotation is passing through its centre is
$I = mR^2$
$\therefore \:\:\:\: I_1 = m_1R^2 $ and $I_2 = m_2 (nR)^2$
Since, both have same density
$\therefore \:\: \frac{m^{2}}{2\pi\left(nR\right)\times A}=\frac{m_{1}}{2\pi R\times A}$
where $A$ is cross-section area of ring.
$\therefore \:\:\: m_2 = nm_1 $
$\because \:\:\: \frac{I_{1}}{I_{2}} = \frac{m_{1}R^{2}}{m_{2}\left(nR\right)^{2} } = \frac{m_{1}R^{2}}{m_{1}n\left(nR\right)^{2}} = \frac{1 }{n^{3}}$
$\because \:\:\: \frac{I_1}{I_2} = \frac{1}{8} $ (Given)
$\therefore \:\:\: \frac{1}{8} = \frac{1}{n^3} $ or $n = 2$