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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 given 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

A

`1//2`

B

`7//9`

C

3

D

7

Text Solution

Verified by Experts

The correct Answer is:
B
Gravitational force due to solid sphere,
`F_(1)=(GMm)/((2R)^(2))`, where M and m are mass of the solid sphere and particle respectively and R is the radius of the sphere. The gravitational force on particle due to sphere with cavity= force due to solid sphere-force due to sphere creating cavity, assumed to be present above at that position.
i.e., `F_(2)=(GMm)/(4R^(2))-(G(M//8)m)/((3R//2)^(2))=(7)/(36)(GM m)/(R^(2))`
So, `(F_(2))/(F_(1))=(7 GMm)/(36R^(2))//((GM m)/(4 R^(2)))=(7)/(9)`
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