A solid sphere of uniform density and radius `R` applies a gravitational force 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 shows 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. `7//9`
C. `3`
D. `7`

Answer

B

Explanation

`C rarr cavity`, `T rarr` Total, `R rarr` remaining
`F_(1) = (GMm)/((2R)^(2))`..(i)
`F_(R ) = F_(2) = F_(T) - F_(C) = F_(1) - F_(C)`
or `F_(2) = (GMm)/((2R)^(2)) - (G((M)/(8)) m)/((3R//2)^(2))`
or `F_(2) = (14 GMm)/(72 R^(2))`
From Eqs. (i) and (ii) we get,
`(F_(2))/(F_(1)) = (7)/(9)`