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, distant 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

Answer

`(7)/(9)`

Explanation

`F_(1)=(GMm)/(4R^(2))`
`F_(2)=` force due to whole sphere - force due to cavity
`F_(2)=(GMm)/(4R^(2))-(GMm)/(18R^(2))implies(7GMm)/(36R^(2))therefore(F_(2))/(F_(1))=(7)/(9)`