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 shown in figure. A particle of mass `m_(0)` is released from rest at centre `B` of the cavity. Calculate velocity with which particle strikes the centre `A` of the sphere. Neglect earth's gravity. Initially sphere and particle are at rest.

Options

A. `sqrt(2/3piGrhoR^(2))`
B. `sqrt(2/3piGrho^(2)R^(2))`
C. `sqrt(2/5piGrhoR^(2))`
D. `sqrt(2/3pi^(2)G^(2)rho^(2)R^(2))`

Answer

A

Explanation

Applying conservation of mechanical energy. Increases in kinetic energy =decreases in gravitational potential energy
or `1/2mv^(2)=U_(B)-U_(A)=m(V_(B)-V_(A))`
`v=sqrt(2(V_(B)-V_(A)).....(i)`
potential at `A: V_(A)` =potential due to complete sphere -potential due to gravity
`=-(1.5GM)/(R)-[-(Gm)/(R//2)]=(2GM)/R`
Here, `m=4/3pi(R/2)^(3)rho=(pirhoR^(3))/6` and `M=4/3piR^(3)rho`
Subtituting the values, we get
`V_(B)=-(GM)/(R^(3))[1.5R^(2)-0.5(R/2)^(2)]+(1.5Gm)/(R//2)`
`=-11/8(GM)/R+(3GM)/R`
`=G/R[(pirhoR^(3))/2-11/6.pirhoR^(3)]=-4/3piGrhoR^(2)`
`V_(B)=V_(A)=1/3piGrhoR^(2)` So, from eq.(1)
`v=sqrt(2/3piGrhoR^(2))`