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(A) Let $\sigma$ be the surface mass density of the discs.Mass of the bigger disc $M = \sigma \pi (2R)^2 = 4 \sigma \pi R^2$.Mass of the removed small

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$\frac{1}{3}$

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(A) Let $\sigma$ be the surface mass density of the discs.Mass of the bigger disc $M = \sigma \pi (2R)^2 = 4 \sigma \pi R^2$.Mass of the removed smaller disc $m = \sigma \pi R^2$.Let the origin be at the centre of the bigger disc. The centre of the removed disc is at a distance $d = R$ from the origin.The centre of mass of the remaining part is given by $X_{cm} = \frac{M(0) - m(d)}{M - m}$.$X_{cm} = \frac{0 - (\sigma \pi R^2)(R)}{4 \sigma \pi R^2 - \sigma \pi R^2} = \frac{-\sigma \pi R^3}{3 \sigma \pi R^2} = -\frac{R}{3}$.The magnitude of the distance is $\frac{R}{3}$,so $\alpha R = \frac{R}{3}$,which gives $\alpha = \frac{1}{3}$.

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

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