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

Options

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

Answer

D

Explanation

The gravitational force due to the whole sphere at A point is
`F_(1) = (GM_(e)m_(0))/((2R)^(2)) ,` where `m_(0)` is the assumed rest mass at point A.
In the second case , when we made a cavity of radius (R/2), then gravitational force at point A is
`F_(2) = (GM_(e)m_(0))/((R + R//2)^(2)) " " therefore F_(2)//F_(1) = 1//9 `