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

Options

A. `1/2`
B. `3/4`
C. `7/8`
D. `9/7`

Answer

D

Explanation

Mass of the cavity `= M//8` if mass of the sphere `=M` as volume of the cavity is `1/8th` of the sphere.
`F_(2)=(GMm)/(4R^(2))-(GMm)/(8(3/2R)^(2))`
`(GMm)/(R^(2))[1/4-1/18]=(GMm)/(R^(2))[(9-2)/36]`
`=(GMm)/(R^(2)) 7/36`
`:. (F_(1))/(F_(2))=36/(4xx7)=9/7`