The gravitational force acting on a particle due to a solid sphere of uniform density and radius R, at a distance of 3R from the center of the sphere is `F_1` . A spherical hole of radius (R/2) is now made in the sphere as shown in the figure . The sphere with hole now exert a force `F_2` on the same particle , Ratio of `F_1 and F_2` is

Options

A. `50/41`
B. `41/50`
C. `41/42`
D. `25/41`

Answer

A

Explanation

Gravitational force due to solid sphere is ,
`F_1= (GMm)/(3R^2) = (GMm)/9R^2`
Gravitational force on particle due to sphere with cavity ,
`F_2= (GMm)/(9R^2) = (G(M/8)m)/(((5)R/2)^2)=(GMm)/R^2[1/9-4/((8xx25)]` = `(GMm)/R^2[4/(50xx9))`
therefore `(F_1)/(F_1)= ((GMm)/(9R^2))/((GMm)/(R^2)[41/((50xx9))])`= 50/41