A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F₁ 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 figure. The sphere with cavity now applies a gravitational force F₂ on same particle placed at P. The ratio F₂/F₁ will be :

7/9

\(F_1 = \frac{G.\frac{4}{3} \pi R^3 \rho m}{4R^2} = \frac{1}{3} \pi \rho G m R\)

\(F_2 = \frac{G.\frac{4}{3} \pi R^3 \rho m}{4R^2} - \frac{G.\frac{4}{3} \pi \frac{R^3}{8} \rho m}{\frac{9}{4}R^2}\)

  = \(\frac{1}{3} \pi \rho G m R - \frac{2}{27} \pi \rho G m R\)

\(\frac{F_1}{F_2} = \frac{\frac{1}{3}}{\frac{1}{3} - \frac{2}{27}}\)

  = \(\frac{1}{1-\frac{2}{9}}\) = \(\frac{9}{7}\)

\(\Rightarrow \frac{F_2}{F_1} = \frac{7}{9}\)