A solid spere of uniform density and radius R applies a gravitational force of attraction equal to `F_(1)` on a particle placed P distacne 2R from the centre O of the spere A spherical cavity of radius `R//2` is now made in the sphere as shown in the The spere with cavity now applies gravitional force `F_(2)` on same particle placed at P the ratio `F_(2)// F_(1)` will be

Options

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

Answer

B

Explanation

Gravitational force due to solid sphere
`F_(1)=(GMm)/(2R)^(2)` where M and m are mass of the solid spere and pariticle respectively and R is the radius of the spere The gravitational force on particle due to spehere with cavity is equal to the gravitatinal foce due to solid sphere creating cavity assumed to be present above at that position
Denisity of spere `=(M)/(4/3 pi R^(3))`
Radius of spere creating cavity `R_(1)=(R )/(2)`
Mass of sphere creating cavity `M^(p)=p.v`
`=(M)/(4/3pi R^(3)).4/3 pi (R)/(2)^(3)=(M)/(8)`
i.e `F_(2)=(GMm)/(4R^(2))-(G(M//8)m)/(3R//2)^(2)=7/36.(GMm)/(R^(2))`
so `(F_(2))/(F_(1))=(7GMm)/(36R^(2))/(GMm)/(4R^(2))=7/9`