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(C) Let the mass per unit area of the plate be $\sigma$.Mass of the original plate $M = \sigma \pi (4r)^2 = 16 \sigma \pi r^2$.Mass of the removed cir

Vedclass · Accepted Answer

$\frac{2 r}{15}$

Vedclass · Answer

(C) Let the mass per unit area of the plate be $\sigma$.Mass of the original plate $M = \sigma \pi (4r)^2 = 16 \sigma \pi r^2$.Mass of the removed circular part $m = \sigma \pi r^2$.The mass of the remaining portion is $M' = M - m = 16 \sigma \pi r^2 - \sigma \pi r^2 = 15 \sigma \pi r^2$.Let the centre of the original plate be at the origin $(0, 0)$.The centre of the removed part is at $(2r, 0)$.Using the principle of centre of mass for a cavity: $M X_{CM} = M' X_{R} + m X_{C}$,where $X_{CM}$ is the centre of mass of the original plate (which is $0$),$X_{R}$ is the centre of mass of the remaining portion,and $X_{C}$ is the centre of mass of the removed part.$0 = (15 \sigma \pi r^2) X_{R} + (\sigma \pi r^2)(2r)$.$15 X_{R} = -2r$.The magnitude of the distance is $|X_{R}| = \frac{2r}{15}$.

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