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 P, distance 2R from the centre O of the sphere. A spherical cavity of radius 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 same particle placed at P. The ratio $F_2/F_1$ will be

• A. 1/2
• B. 7/9
• C. 3
• D. 7

Answer 1

The correct option is B 7/9

Let us assume the mass of the sphere as $M$

First, find out the force by the intact sphere by $F1=\frac{GMm}{4R^2}$

To find $F2$ divide the cavity sphere into $2$ sphere of positive mass and a negative mass like we do in all cavity problems.

Find out the mass of negative sphere $=M/B$ ( as the density is same mass will depend on the volume)

after this we find out the forces by these $2$ sphere of masses $M,-M/B$ at a distance of $2R$ $3R/2$ respectively and add them

$F_{removed-mass}=\frac{G\frac{-M}{8}m}{\frac{(3R{)}^2}{2}}$

$F_2=\frac{GMm}{18R^2}$

Where, $M$ is the mass of the spere and $m$ is the mass of the particle.

So, $F_2=\frac{GMm}{4R^2}-\frac{GMm}{18R^2}=\frac{7GMm}{36R^2}$

find $\frac{F_2}{F_1}=\frac{7}{9}$