Question

A small ball of relative density $0.8$ falls into water from a height of $2m$. The depth to which, the ball will sink, is (neglect viscous forces)

• A. $8m$
• B. $2m$
• C. $6m$
• D. $4m$

Answer 1

The correct option is A $8m$

Velocity of the ball as it reaches water surface is $\sqrt{2gh}$ i.e. $\sqrt{4g}$
Lets say, $\rho$ is the density of ball and $\sigma$ is the density of water. Now
Net upward force acting on the body $F=\sigma Vg-\rho Vg$
$\Rightarrow ma=\sigma Vg-\rho Vg$
$\Rightarrow \left(\rho V\right)a=Vg(\sigma -\rho )$
$\Rightarrow a=g(\frac{\sigma }{\rho }-1)=g(\frac{1}{0.8}-1)=\frac{g}{4}$ in upward direction
lets say it sinks to the depth $H$, at this instant final velocity becomes $v=0$, so we have
$0=(\sqrt{4g}{)}^2-2\times \frac{g}{4}\times H$ $\because v^2=u^2+2as$
$4g=\frac{2g}{4}H\Rightarrow H=8m$