Inside a fixed sphere of radius `R` and uniform density `rho`, there is spherical cavity of radius `(R )/(2)` such that surface of the cavity passes through the centre of the sphere as shows in figure. A particle of mass `m_(0)` is released from rest at centre `B` of the cavity. Caculate velocity with which particle strikes the centre `A` of the sphere. Neglect earth's gravity. Initially sphere and particle are at rest.

Answer

B, C

Explanation

Applying conservation of machanical energy,
Increase in kinetic energy
`=` decrease in gravitational potential energy
or `(1)/(2) m_(0)nu^(2) = U_(B) - U_(A) = m_(o) (V_(B) - V_(A))`
`:. nu = sqrt(2(V_(B) - V_(A))` ..(i)
Potential at `A`
`V_(A) =` potential due to complete sphere
- potential due to cavity
`= - (1.5 GM)/(R ) - [-(GM)/(R//2)]`
`= (2Gm)/(R ) - (1.5 GM)/(R )`
Hence, `m = (3)/(4) pi ((R )/(2))^(3) rho = (pi rho R^(3))/(6)`
and `M = (4)/(3) pi R^(3) rho`
Substituting the value, we get
`V_(A) = (G)/(R ) [(pi rho R^(3))/(3) - 2 pi rho R^(3)] = - (5)/(3) pi G rho R^(2)`
Potential at `B`
`V_(B) = (GM)/(R^(3))[1.5 R^(2) - 0.5((R )/(2))^(2)] + (1.5 Gm)/(R//2)`
`= - (8)/(11) (GM)/(R ) + (3 Gm)/(R )`
`=(G)/(R)[(pi rho R^(3))/(2) - (11)/(6). pi rho R^(3)] = - (4)/(3) pi G rho R^(2)`
`:. V_(B) - V_(A) = (1)/(3) pi G rho R^(2)`
So, from Eq. (i)
`nu = sqrt((2)/(3) pi G rho R^(2))`