A small body of superdense material, whose mass is twice the mass of the earth but whose size is very small compared to the size of the earth, starts form rest at a height `H lt lt R` above the earth's surface, and reaches the earth's surface in time `t`. then `t` is equal to

Options

A. `t=sqrt((h)/(g))`
B. `t=sqrt((2h)/(g))`
C. `t=sqrt((2h)/(3g))`
D. `t=sqrt((4h)/(3g))`

Answer

C

Explanation

Acceleration of super dense material = g
But acceleration of earth = 2g (towards the mass)
`:.` Relative acceleration = 3g
Since, we know that, `h = ut + (1)/(2)a_(r)t^(2)`
But the body starts from rest
So, `u=0`
Now, `h=(1)/(2)a_(r)t^(2)or h=(1)/(2)(3g)t^(2)or t=sqrt((2h)/(3g))`.