Ball Fall

Net force acting on the ball when it enters the liquid

$\sigma.V.g$ the buoyant force upwards

And

$\rho.V.g$ the weight of the ball downwards

Here $V$ is the volume of the ball

Let's assume it moves a distance $X$ down in the water.

Applying energy conservation from the point it enters the water to the point it becomes stationary in the water for an instant.

Loss in kinetic energy $=$ Work done against the net force

$\frac {m.v^2}{2} = F.X$

$\frac {\rho. V.2.g.H}{2} = (\sigma.V.g - \rho.V.g).X$

$V$ and $g$ gets cancelled and by rearranging the terms we can find the answer

I am actually getting the answer as h\quad =\quad \frac { \rho }{ \sigma -2\rho } H. This is because the fall in potential energy = work done against buoyant force. Taking the depth in water to which it goes as h, the equation then is V\times \rho \times g\times (H+h)\quad =\quad V\times (\sigma -\rho )\times g\times h which gives the final answer as h\quad =\quad \frac { \rho }{ \sigma -2\rho } H.

Step 1: When ball touch the water level then it stored total energy= mgh =roh *g * H

Step 2: let Total energy consume when ball sink to a height h, then Total energy Consume= (sigma- roh)* g *h

Step 3: Total energy stored= Total energy Consumed then roh g * H= (Sigma- roh) * g h

Step 4: h= (roh*H)/(sigma-roh)