Consider two uniform discs of the same thickness and different radii $R_{1}=R$ and $R_{2}=\alpha R$ made of the same material. If the ratio of their moments of inertia $I_{1}$ and $I_{2}$, respectively, about their axes is $I_{1}: I_{2}=1: 16$ then the value of $\alpha$ is:
Correct Option: , 3
(3) Let $p$ be the density of the discs and $t$ is the thickness of discs.
Moment of inertia of disc is given by
$I=\frac{M R^{2}}{2}=\frac{\left[\rho\left(\pi R^{2}\right) t\right] R^{2}}{2}$
$I \propto R^{4}$ (As $\rho$ and $t$ are same)
$\frac{I_{2}}{I_{1}}=\left(\frac{R_{2}}{R_{1}}\right)^{4} \Rightarrow \frac{16}{1}=\alpha^{4} \Rightarrow \alpha=2$
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