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(C) Let $\rho$ be the uniform density of the sphere.Mass of the original sphere of radius $R$ is $M = \frac{4}{3} \pi R^{3} \rho$.Mass of the spherica

Vedclass · Accepted Answer

$\left(R^{2}+R+1\right)(2-R)=1$

Vedclass · Answer

(C) Let $ho$ be the uniform density of the sphere.Mass of the original sphere of radius $R$ is $M = \frac{4}{3} \pi R^{3} ho$.Mass of the spherical cavity of radius $1$ is $m = \frac{4}{3} \pi (1)^{3} ho = \frac{4}{3} \pi ho$.The mass of the remaining part is $M' = M - m = \frac{4}{3} \pi ho (R^{3} - 1)$.The centre of mass of the original sphere is at $C$. The centre of mass of the cavity is at $O$. The distance $CO = R - 1$.The centre of mass of the remaining part is at $G$,which is on the surface of the cavity,so $CG = R - 1$ (distance from $C$ to the edge of the cavity).Using the principle of moments about the centre of mass $C$:$M' 	imes CG = m 	imes CO$$\left[\frac{4}{3} \pi ho (R^{3} - 1)ight] 	imes (R - 1) = \left[\frac{4}{3} \pi hoight] 	imes (R - 1)$Since $R 
eq 1$,we can divide by $(R - 1)$:$R^{3} - 1 = 1$Wait,let's re-evaluate the distance $CG$. From the figure,$G$ is the centre of mass of the remaining part. The distance of $G$ from $C$ is $x$. The distance of $O$ from $C$ is $R-1$. The distance of $G$ from $O$ is $1$. Thus,$CG = CO - OG = (R-1) - 1 = R-2$. Since $G$ is to the left of $C$,we take the magnitude as $2-R$.Using $M' 	imes CG = m 	imes CO$:$\frac{4}{3} \pi ho (R^{3} - 1) 	imes (2 - R) = \frac{4}{3} \pi ho 	imes (R - 1)$$(R^{3} - 1)(2 - R) = R - 1$$(R - 1)(R^{2} + R + 1)(2 - R) = (R - 1)$$(R^{2} + R + 1)(2 - R) = 1$

As shown in the figure,when a spherical cavity (centred at $O$) of radius $1$ is cut out of a uniform sphere of radius $R$ (centred at $C$),the centre of mass of the remaining (shaded) part of the sphere is at $G$,i.e.,on the surface of the cavity. $R$ can be determined by the equation:

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