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(B) Let $\rho$ be the density of the lead sphere.Let $M$ be the mass of the original sphere of radius $R$. Then,$M = \frac{4}{3} \pi R^3 \rho$.The hol

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$\frac{R}{14}$

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(B) Let $ho$ be the density of the lead sphere.Let $M$ be the mass of the original sphere of radius $R$. Then,$M = \frac{4}{3} \pi R^3 ho$.The hollow sphere has a radius $r = \frac{R}{2}$.The mass of the removed part is $m_1 = \frac{4}{3} \pi (\frac{R}{2})^3 ho = \frac{M}{8}$.The mass of the remaining part is $m_2 = M - m_1 = M - \frac{M}{8} = \frac{7M}{8}$.Let the centre of the original sphere be the origin $(0,0)$.The centre of the removed part is at $x_1 = \frac{R}{2}$.Let the centre of mass of the remaining part be at $x_2$.Since the centre of mass of the original sphere was at the origin,we have:$X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = 0$$0 = \frac{(\frac{M}{8})(\frac{R}{2}) + (\frac{7M}{8})(x_2)}{M}$$0 = \frac{MR}{16} + \frac{7M}{8} x_2$$x_2 = -\frac{MR}{16} 	imes \frac{8}{7M} = -\frac{R}{14}$.The negative sign indicates that the centre of mass shifts by $\frac{R}{14}$ in the direction opposite to the cavity.

$A$ 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 centre. What is the shift in the centre of mass of the lead sphere due to this?

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