A spherical hollow is made in a lead sphere of radius `R` such that its surface touches the outside surface of the lead sphere and passes through the centre. The mass of the lead sphere before hollowing was `M`. The force of attraction that this sphere would exert on a particle of mass `m` which lies at a distance `d` from the centre joining the centre of the sphere and the hollow is

Options

A. `(GMm)/(d^(2))`
B. `(GMm)/(d^(2))[1-1/(8(1-R/(2d))^(2))]`
C. `(GMm)/(d^(2))[1-1/(8(1+R/(2d))^(2))]`
D. `(GMm)/(8d^(2))`

Answer

B

Explanation

`F_(1)=`gravitational force between sphere of mass `M` and the particle of mass `m=(GMm)//d^(2)`.
`F_(2)`=gravitational force between the removed part and the particle of mass `m=(GM_(1)m)//(d-R//2)^(2)`
where `M_(1)`= mass of the removed part
`=4/3pi(R/2)^(3)rho=(4piR^(3)rho)/(3xx8)=M/8`
`:. F_(2)=(GM_(1)m)/(8(d-R/2)^(2))`
where `M_(1)`=mass of the removed part
`=4/3pi(R/2)^(2)rho=(rpiR^(3)rho)/(3xx8)=M/8` So, `F_(2)=(GMm)/(8(d-R/2)^(2))`
Hence, required force
`F=F_(1)-F_(2)=(GMm)/(d^(2))-(GMm)/(8(d-R/2)^(2))`
`=(GMm)/(d^(2))[1-1/(8(1-R/(2d))^(2))]`