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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

Text Solution

Verified by Experts

The correct Answer is:
`(7)/(9)`
`F_(1)=(GMm)/(4R^(2))`
`F_(2)=` force due to whole sphere - force due to cavity
`F_(2)=(GMm)/(4R^(2))-(GMm)/(18R^(2))implies(7GMm)/(36R^(2))therefore(F_(2))/(F_(1))=(7)/(9)`
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