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A circular disc of radius $\mathrm{R}$ is removed from a bigger circular disc of radius $\mathrm{2R}$ such that the circumferences of the discs coincide. The centre of mass of the new disc is $\mathrm{\alpha / R}$ from the centre of the bigger disc. The value of $\mathrm{\alpha}$ is
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The correct answer is:
$1 / 3$
$1 / 3$
In this question distance of centre of mass of new disc is $\alpha R$ not $\frac{\alpha}{R}$.
$\begin{aligned}
& -\frac{3 \mathrm{M}}{4} \alpha \mathrm{R}+\frac{\mathrm{M}}{4} \mathrm{R}=0 \\
& \Rightarrow \alpha=\frac{1}{3}
\end{aligned}$
$\begin{aligned}
& -\frac{3 \mathrm{M}}{4} \alpha \mathrm{R}+\frac{\mathrm{M}}{4} \mathrm{R}=0 \\
& \Rightarrow \alpha=\frac{1}{3}
\end{aligned}$
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