Q.
A spherical hollow 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. What is the shift in the centre of lead sphere as a result of this hollowing?
A) $\frac{R}{7}$
B) $\frac{R}{14}$
C) $\frac{R}{2}$
D) $R$
Explanation
(b) Let centre of mass of lead sphere after hollowing be at point $O_{2},$ where $OO_{2} = x$
Mass of spherical hollow $m = \frac{\frac{4}{3}\pi\left( \frac{R}{2} \right)^{2}M}{\left( \frac{4}{2}\pi R^{3} \right)} = \frac{M}{8}$ and
$x = OO_{1} = \frac{R}{2}$
[image:7_image102.emf]
$\therefore\ x = \frac{M \times 0 - \left( \frac{M}{8} \right) \times \frac{R}{2}}{M - \frac{M}{8}} = \frac{\frac{MR}{16}}{\frac{7M}{8}} = - \frac{R}{14}$
$\therefore$ shift $= \frac{R}{14}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.