A diametrical tunnel is dug across the earth. A ball dropped into the tunnel from one side. The velocity of the ball when it reaches the centre of the earth is [Given: gravitational potential at the centre of earth `= -3//(2GM//R)`]

Options

A. `sqrt(R)`
B. `sqrt(gR)`
C. `sqrt(2.5gR)`
D. `sqrt(7.1gR)`

Answer

B

Explanation

Gravitational potential at a point on the surface of earth is `-(GM)/R`
If the earth is assumed to be a solid sphere, then the gravitational potential at the centre of the earth is `(3/(2GM)//R)`.
Decrease in gravitational potential is
`R/2xx(GM)/(R^(2))=(Rg)/2`
Loss is potential energy is
`R/2xx(GM)/(R^(2))xxm`
Now gain in kinetic energy `=` loss is potnetial energy. Therefore.
`1/2mv^(2)=1/2mgR` or `v=sqrt(gR)`