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(B) Let the mass of the entire disc be $M$.The mass per unit area is $\sigma = \frac{M}{\pi(2R)^2} = \frac{M}{4\pi R^2}$.The mass of the removed circu

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$\frac{1}{3}$

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(B) Let the mass of the entire disc be $M$.The mass per unit area is $\sigma = \frac{M}{\pi(2R)^2} = \frac{M}{4\pi R^2}$.The mass of the removed circular disc of radius $R$ is $M_1 = \sigma \cdot \pi R^2 = \frac{M}{4\pi R^2} \cdot \pi R^2 = \frac{M}{4}$.The mass of the remaining disc is $M_2 = M - M_1 = M - \frac{M}{4} = \frac{3M}{4}$.Let the centre of the bigger disc be the origin $(0,0)$. The centre of mass of the removed disc is at $x_1 = R$ and the centre of mass of the remaining disc is at $x_2 = -\alpha R$.Since the centre of mass of the original disc was at the origin,we have:$M_1 x_1 + M_2 x_2 = 0$$\frac{M}{4} \cdot R + \frac{3M}{4} \cdot (-\alpha R) = 0$$\frac{M}{4} \cdot R = \frac{3M}{4} \cdot \alpha R$$\alpha = \frac{1}{3}$.

$A$ circular disc of radius $R$ is removed from one end of a bigger circular disc of radius $2R$. 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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