Question

A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to $F_1$ on a particle placed at a distance 3R from the centre of the sphere. A spherical cavity of radius $\frac{R}{2}$ is now made in the sphere as shown in the figure. The sphere with cavity now applies a gravitational force $F_2$ on the same particle. The ratio $\frac{F_2}{F_1}$ is:

• A. $\frac{9}{50}$
• B. $\frac{41}{50}$
• C. $\frac{3}{25}$
• D. $\frac{22}{25}$

Answer 1

The correct option is B $\frac{41}{50}$

$F_1=\frac{GMm}{9R^2}where,M=\rho \times \frac{4}{3}\pi R^3$
When a sphere of radius $\frac{R}{2}$ is removed
$M^1=\rho \frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3=\frac{M}{8}$
The new force is,
$F_2=\frac{GMn}{R^2}-\frac{G\left(\frac{M}{8}\right)m}{\left(\frac{25R^2}{4}\right)}$
$=\frac{GMm}{9R^2}[\frac{1}{9}-\frac{1}{50}]$
$=\frac{GMm}{R^2}\left[\frac{41}{9\left(50\right)}\right]$
$\frac{F_2}{F_1}=\frac{41}{50}$