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A solid sphere of uniform density and ra...

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

`(F)/(3)`

B

`(2F)/(3)`

C

`(4F)/(3)`

D

`(7F)/(9)`

Text Solution

Verified by Experts

The correct Answer is:
D
Gravitational force of attraction on mass `m` at `P` due to soild sphere is
`F = (GM m)/((2R)^(2)) = (GM m)/(4R^(2))` or `(GM m)/(R^(2)) = 4F` ..(i)
Mass of the spherical portion removed from sphere
`M' = (M)/((4)/(3) piR^(3)) xx (4)/(3) pi (R//2)^(3) = (M)/(8)`
Gravitational force of attraction on mass `m` at `P` if mass of the spherical portion removed is present there is
`F' = (G(M//8)m)/((3R//2)^(2)) = (GM m)/(R^(2)) xx (1)/(8) xx (4)/(9)`
`= 4F xx (1)/(8) xx (4)/(9) = (2F)/(9)`
`:.` Gravitational force of attraction on mass `m` at `P` due to remaining part of the sphere is
`F'' = F - F' = F - (2F)/(9) = (7F)/(9)`
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