Question

A solid sphere of uniform density and radius R applied a gravitational force of attraction equal to $F_1$ on a particle placed at a distance $2R$ from the centre of the sphere. A spherical cavity of radius $R/2$ is now made in the sphere as shown in the figure. The sphere with the cavity now applied a gravitational force $F_2$ on the same particles. The ratio $F_1/F_2$ is?

• A. $\frac{1}{2}$
• B. $\frac{3}{4}$
• C. $\frac{7}{8}$
• D. $\frac{9}{7}$

Answer 1

The correct option is D $\frac{9}{7}$

$F_1=P\frac{4}{3}\frac{\pi R^3}{(2R{)}^2}$

$F_2=P\frac{4}{3}\pi \frac{R^3}{(2R{)}^2}-P\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3{\left(\frac{2}{3R}\right)}^2$

$=P\frac{4}{3}\pi \frac{7}{36}R$

$\Rightarrow \frac{F_1}{F_2}=\frac{1}{4}\times \frac{36}{7}=\frac{9}{7}$