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(B) Let the mass per unit area be $\sigma$.The mass of the complete disc of radius $2R$ is $M_1 = \sigma \pi (2R)^2 = 4\pi \sigma R^2$. Its center of

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$\frac{1}{3}$

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(B) Let the mass per unit area be $\sigma$.The mass of the complete disc of radius $2R$ is $M_1 = \sigma \pi (2R)^2 = 4\pi \sigma R^2$. Its center of mass is at the origin $O$.The mass of the removed disc of radius $R$ is $M_2 = \sigma \pi R^2$. Its center of mass is at a distance $R$ from the origin $O$ along the positive x-axis.Using the concept of negative mass for the cavity,the center of mass of the remaining part is given by:$x_{cm} = \frac{M_1 x_1 - M_2 x_2}{M_1 - M_2}$$x_{cm} = \frac{(4\pi \sigma R^2)(0) - (\pi \sigma R^2)(R)}{4\pi \sigma R^2 - \pi \sigma R^2}$$x_{cm} = \frac{-\pi \sigma R^3}{3\pi \sigma 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 circular disc of radius $2R$ such that the circumferences of the discs coincide. The center of mass of the new disc is $\alpha R$ from the center of the bigger disc. The value of $\alpha$ is:

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