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(D) Let the radius of the original disc be $R = 20 \ cm$ and the radius of the hole be $r = 5 \ cm$.The mass of the original disc is $M = \sigma \pi R

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$1.0$

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(D) Let the radius of the original disc be $R = 20 \ cm$ and the radius of the hole be $r = 5 \ cm$.The mass of the original disc is $M = \sigma \pi R^2$,where $\sigma$ is the surface mass density.The mass of the cut-out part is $m = \sigma \pi r^2$.Since $r = R/4$,the mass $m = \sigma \pi (R/4)^2 = M/16$.The centre of mass of the original disc is at the origin $(0, 0)$.The centre of the hole is at a distance $d = R - r = 20 - 5 = 15 \ cm$ from the origin.Let the centre of mass of the remaining part be $X_{	ext{com}}$. Using the formula for the centre of mass of a system with a cavity:$X_{	ext{com}} = \frac{M(0) - m(d)}{M - m}$$X_{	ext{com}} = \frac{0 - (M/16)(15)}{M - M/16} = \frac{-(15/16)M}{(15/16)M} = -1 \ cm$.The magnitude of the distance is $|X_{	ext{com}}| = 1 \ cm$.

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