A body of relative density `5.0` is released from rest on the surface water filled to a height of `1.0` m in a tall cylinder. If viscous force is neglected, find the time taken by the body to reach the bottom of the cylinder. Take `g=10ms^(-2)`.

Options

A. `0.25`
B. `0.5`
C. `0.125`
D. `1`

Answer

B

Explanation

Density of body `(rho)=5000kgm^(-3)`
Density of water `(sigma)=1000kg m^(-3)`
If V is the volume of the body, its mass `(m)=Vrho`.
Weight of the body `(W)=mg=Vrhog`
Upthrust on the body (U)`=Vsigmag :. `Net downward force `F=W-U=(rho-sigma)Vg`
`:.` Acceleration `a=(F)/(m)=(F)/(rhoV)=((rho-sigma)Vg)/(rhoV)=((rho-sigma)/(rho))g`
From the relation `s=ut+(1)/(2)at^(2)`, we have `h=0+(1)/(2)at^(2)`
`impliest=sqrt((2h)/(a))=[(2hrho)/((rho-sigma)g)]^(1//2)=((2xx1.0xx5000)/((5000-1000)xx10))^(1//2)=0.5s`