A spherical cavity is made in a lead sphere of radius R such that its surface touches the outside surface of…

A spherical cavity is made in a lead sphere of radius $\mathrm{R}$ 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 of this hollowing, is

1. $\frac{\text{R}}{7}$
2. $\frac{\text{R}}{14}$
3. $\frac{\text{R}}{2}$
4. R

Solution

Let centre of mass of lead sphere after hollowing be at point ${\mathrm{O}}_2$, where ${\mathrm{OO}}_2=\mathrm{x}$

Mass of spherical hollow $\mathrm{m}=\frac{\frac{4}{3}\mathrm{\pi }{\left(\frac{\text{R}}{2}\right)}^2\text{M}}{\left(\frac{4}{2}\mathrm{\pi }{\text{R}}^3\right)}=\frac{\text{M}}{8}$ and

$\text{x}={\text{OO}}_1=\frac{\text{R}}{2}$



$\therefore \text{x}=\frac{\text{M}\times 0-\left(\frac{\text{M}}{8}\right)\times \frac{\text{R}}{2}}{\text{M}-\frac{\text{M}}{8}}=\frac{\frac{\text{MR}}{16}}{\frac{7\text{M}}{8}}=-\frac{\text{R}}{14}$

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

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