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(B) Let the centre of the bigger disc be the origin $O(0,0)$.Let the mass of the original disc be $M$ and its radius be $2R$.The mass of the removed s

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$\frac{1}{3}$

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(B) Let the centre of the bigger disc be the origin $O(0,0)$.Let the mass of the original disc be $M$ and its radius be $2R$.The mass of the removed smaller disc of radius $R$ is $m = \frac{\pi R^2}{\pi (2R)^2} M = \frac{M}{4}$.The centre of the removed disc is at a distance $R$ from the origin along the $x$-axis,so its coordinates are $(R, 0)$.The centre of mass of the remaining part $(x_{CM})$ is given by the formula:$x_{CM} = \frac{M_1 x_1 - m x_2}{M_1 - m}$Here,$M_1 = M$,$x_1 = 0$,$m = M/4$,and $x_2 = R$.$x_{CM} = \frac{M(0) - (M/4)(R)}{M - M/4} = \frac{-MR/4}{3M/4} = -\frac{R}{3}$.The distance from the centre is $|x_{CM}| = \frac{R}{3}$.Comparing this with $\alpha R$,we get $\alpha = \frac{1}{3}$.

$A$ circular disc of radius $R$ is removed from a bigger circular disc of radius $2R$ such that the circumferences of the discs touch. The centre of mass of the new disc is at a distance $\alpha R$ from the centre of the bigger disc. The value of $\alpha$ is

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