Inside a fixed sphere of radius R and uniform density ρ, there is spherical cavity of radius R2 such that surface of the cavity passes through the centre of the sphere as shown in figure. A particle of mass m is released from rest at centre B of the cavity. The velocity with which particle strikes the centre A of the sphere is v. Neglect earth's gravity. Initially sphere and particle are at rest. If the escape speed of the particle from surface of the outer bigger sphere is ve. Then find 2ve2v2.

Applying conservation of mechanical energy, we have     (U+K)at B=(U+K)at A⇒UB+KB=UA+KA⇒UB+0=UA+12mvA2where UB=mVB and UA=mVA, so we have    12mvA2=UB−UA=mVB−VA⇒vA=2VB−VA     ........(1)Potential at A        VA= Potential Due  to the Complete  Sphere at A− Potential Due  to the Spherical  Cavity at AVA=−32GMR−−GM′r=GM′r−32GMRwhere        M=43πR3ρ,r=R2 and M′=43πr3ρ=πρR36Substituting the values, we get       VA=GRπρR33−2πρR3=−53πGρR2Potential at B         VB= Potential Due  to the Complete  Sphere at B− Potential Due  to the Spherical  Cavity at B⇒ VB=−GM2R33R2−r2−−32GM′rwhere     M=43πR3ρ,r=R2 and M′=43πr3ρ=πρR36    ⇒VB=−118GMR+3GM′R⇒VB=GRπρR32−11πρR36=−43πGρR2⇒VB−VA=13πGρR2So, from equation (1)v=23πGρR2Also, ve=2GMnet Rwhere Mnet =M−M′=76πR3ρ⇒ ve=2G76πR3ρR⇒ ve=73πGρR2⇒ vev=72⇒ ve2v2=72⇒ 2ve2v2=7