A tunnel is dug inside the earth across one of its diameters. Radius of earth is R and its mass is M. A particle is projected inside the tunnel with velocity `sqrt((2GM)/(R))` from one of its ends then maximum velocit attained by the particle in the subsequent motion is `sqrt((nGM)/(R))` (assuming tunnel to be frictionless). Find n

Answer

3

Explanation

Maximum velocity will be at its centre from energy convervation
`(1)/(2)m((2GM)/(R))+m((-GM)/(R))=(1)/(2)mv_(1)^(2)+m((-3)/(2)(GM)/(R))`
`(1)/(2)mv_(1)^(2)=(3)/(2)(GMm)/(R),V_(1)=sqrt((3GM)/(R))`