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(B) Let the centre of the bigger disc of radius $2R$ be at the origin $(0,0)$.Let $M$ be the mass of the bigger disc. The mass per unit area is $\sigm

Vedclass · Accepted Answer

$\frac{1}{3}$

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(B) Let the centre of the bigger disc of radius $2R$ be at the origin $(0,0)$.Let $M$ be the mass of the bigger disc. The mass per unit area is $\sigma = \frac{M}{\pi(2R)^2} = \frac{M}{4\pi R^2}$.The mass of the removed smaller disc of radius $R$ is $m = \sigma \cdot \pi R^2 = \frac{M}{4\pi R^2} \cdot \pi R^2 = \frac{M}{4}$.The centre of mass of the bigger disc is at $x_1 = 0$. The centre of mass of the removed smaller disc is at $x_2 = R$ (since the circumferences touch).The centre of mass of the remaining part is given by:$x_{CM} = \frac{M x_1 - m x_2}{M - m}$$x_{CM} = \frac{M(0) - (M/4)(R)}{M - M/4} = \frac{-MR/4}{3M/4} = -\frac{R}{3}$.The distance from the centre of the bigger disc 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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