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 figure. The particle with cavity now applies a gravitational force `F_(2)` on same particle placed at `P`. The radio `F_(2)//F_(1)` will be

Explanation

`F_1 = (GmM)/((2R)^2) = (GmM)/(4R^2)`
` F_2 = (GmM)/((2R)^2) - (G(M//8)m)/((3R//2)^2) = (7Gm M)/(36R^2)`
` therefore F_2/F_1 =(7GmM)/(36R^2) xx (4R^2)/(GmM) = 7/9`