A small solid ball is dropped from a height above the free surface of a liquid. It strikes the surface of the liqiud at `t=0`. The density of the material of the ball is `500kg//m^(3)` and that of liquid is `1000 kg//m^(3)`. If the ball comes momentarily at rest at `t=2 sec` then initial height of ball form surface of liquid was (neglect viscosity):

Options

A. 20m
B. 10m
C. 15m
D. 25m

Answer

A

Explanation

Velocity of ball when it reaches to surface of liquid
`a = (1000gV-500gV)/(500V)`, where `V` is the volume of the ball
`a = 10 m//sec^(2)`
Apply `v = u+at implies 0 = sqrt(2gh) - 10t`
`implies sqrt(2gh) = 10xx(2)`
`implies 2 xx 10 xx h = 400 implies 20 m`.