Question

In the figure shown, a spherical part of radius $\frac{R}{2}$ is removed from a bigger solid sphere of radius $R$. Assuming uniform mass distribution, shift in the centre of mass will be:

• A. $\frac{R}{7}$
• B. $\frac{R}{14}$
• C. $\frac{R}{9}$
• D. $\frac{R}{6}$

Answer 1

The correct option is D $\frac{R}{6}$

Let the centre of the big sphere which is its centre 
of mass be the origin O. Then the centre of mass 
of the small sphere is at a distance R/2 from O.
When the small sphere is cut out, let the C.M. of the 
remaining portion shifts to P. Mass of remaining portion = 3M/4.
From conservation of centre of mass : 
C.M. of remaining portion = C.M. of big sphere + C.M. of the small sphere.

$\begin{array}{c}\Rightarrow \frac{3M}{4}\times \left(-OP\right)=M\times OO+\frac{M}{4}\times \frac{R}{2}\\ \Rightarrow OP=-\frac{R}{6}\\ so,thecenterofmassofremainingportionshiftsto\frac{R}{6}\\ fromcentreofthecircle.\end{array}$