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 velocity attained by the particle in the subsequent motion is (assuming tunnel to be frictionless)

Options

A. `sqrt((GM)/(R))`
B. `sqrt((3GM)/(R))`
C. `sqrt((2GM)/(R))`
D. `sqrt((5GM)/(R))`

Answer

B

Explanation

maximum velocity will be at its centre. From energy conservation
`(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))`