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Q.

A spherical cavity is made in a lead sphere of radius $R$ such that its surface touches the outside surface of lead sphere and passes through the centre. The shift in the centre of mass of the lead sphere as a result as a result of this hollowing, is

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a

$\frac{R}{14}$

b

$\frac{R}{7}$

c

$R$

d

$\frac{R}{2}$

answer is B.

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Detailed Solution

Let centre of mass of lead sphere hollowing be at point $O_{2,}$ where $O O_{2} = x$
Mass of spherical hollow $m = \frac{\frac{4}{3} π {(\frac{R}{2})}^{2} M}{(\frac{4}{3} π R^{3})} = \frac{M}{8}$ and
$x = O O_{1} = \frac{R}{2} ∴ x = \frac{M \times 0 - (\frac{M}{8}) \times \frac{R}{3}}{M - \frac{M}{8}} = - \frac{R}{14}$

shift $= \frac{R}{14}$

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