Question

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

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

Answer 1

Correct Answer - D
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`