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(B) Let $\rho$ be the density of the lead sphere.The mass of the original sphere is $M = \frac{4}{3} \pi R^3 \rho$.The hollowed part is a sphere of ra

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$\frac{R}{14}$

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(B) Let $\rho$ be the density of the lead sphere.The mass of the original sphere is $M = \frac{4}{3} \pi R^3 \rho$.The hollowed part is a sphere of radius $r = \frac{R}{2}$. Its mass is $m_1 = \frac{4}{3} \pi (\frac{R}{2})^3 \rho = \frac{1}{8} (\frac{4}{3} \pi R^3 \rho) = \frac{M}{8}$.The remaining mass is $m_2 = M - m_1 = M - \frac{M}{8} = \frac{7M}{8}$.Let the center of the original sphere be the origin $(0,0)$. The center of the removed sphere is at $x_1 = \frac{R}{2}$.Using the center of mass formula for the original sphere: $X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = 0$.Substituting the values: $0 = \frac{(\frac{M}{8})(\frac{R}{2}) + (\frac{7M}{8})(x_2)}{M}$.Solving for $x_2$: $\frac{MR}{16} + \frac{7M}{8} x_2 = 0 \implies \frac{7}{8} x_2 = -\frac{R}{16} \implies x_2 = -\frac{R}{14}$.The negative sign indicates the center of mass shifts to the left by $\frac{R}{14}$.

$A$ spherical hollow is made in a lead sphere of radius $R$,such that its surface touches the outside surface of the lead sphere and passes through the center. What is the shift in the center of mass of the lead sphere due to the hollowing?

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