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A circular disc $X$ of radius $R$ is made from an iron plate of thickness $t$, and another disc $Y$ of radius $4 \mathrm{R}$ is made from an iron plate of thickness $\frac{\mathrm{t}}{4}$. Then the relation between the moment of inertia $\mathrm{I}_{\mathrm{X}}$ and $\mathrm{I}_{\mathrm{Y}}$ is
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The correct answer is:
$\mathrm{I}_{\mathrm{Y}}=64 \mathrm{I}_{\mathrm{X}}$
$\mathrm{I}_{\mathrm{Y}}=64 \mathrm{I}_{\mathrm{X}}$
$$
\mathrm{I}=\frac{1}{2} \mathrm{mR} \quad \text { or } \quad \mathrm{M} \propto \mathrm{t} \propto \mathrm{R}^2
$$
For $\operatorname{disc} X, I_x=\frac{1}{2}(m)(R)^2=\frac{1}{2}\left(\pi r^2 t\right) .(R)^2$
For disc $Y, I_y=\frac{1}{2}\left[\pi(4 R)^2 \cdot t / 4\right][4 R]^2$
$$
\Rightarrow \frac{I_x}{I_y}=\frac{1}{(4)^3} \Rightarrow I_y=64 I_x
$$
\mathrm{I}=\frac{1}{2} \mathrm{mR} \quad \text { or } \quad \mathrm{M} \propto \mathrm{t} \propto \mathrm{R}^2
$$
For $\operatorname{disc} X, I_x=\frac{1}{2}(m)(R)^2=\frac{1}{2}\left(\pi r^2 t\right) .(R)^2$
For disc $Y, I_y=\frac{1}{2}\left[\pi(4 R)^2 \cdot t / 4\right][4 R]^2$
$$
\Rightarrow \frac{I_x}{I_y}=\frac{1}{(4)^3} \Rightarrow I_y=64 I_x
$$
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