Question

A small body of density ${\rho }^{\prime }$ is dropped from rest at a height h into a lake of density $\rho$, where $\rho >{\rho }^{\prime }$. Which of the following statements is/are correct if all dissipative effects are neglected (neglect viscosity)?

• A. The speed of the body just entering the lake is $\sqrt{2gh}$.
• B. The body in the lake experiences upward acceleration equal to $\left\{\rho /{\rho }^{\prime }\right)-1\}g$.
• C. The maximum depth to which the body sinks in the lake is $h{\rho }^{\prime }/(\rho -{\rho }^{\prime })$.
• D. The body does not come back to the surface of the lake.

Answer 1

Answer: (A), (B), (C)

The body in the lake experiences upward acceleration equal to $\left\{\rho /{\rho }^{\prime }\right)-1\}g$.
The maximum depth to which the body sinks in the lake is $h{\rho }^{\prime }/(\rho -{\rho }^{\prime })$.
The speed of the body just entering the lake is $\sqrt{2gh}$.

Velocity of the ball as it reaches water surface is $\sqrt{2gh}$ $\Rightarrow$ option A is correct
Net upward force acting on the body $F=\rho Vg-{\rho }^{\prime }Vg$
$\Rightarrow ma=\rho Vg-{\rho }^{\prime }Vg$
$\Rightarrow \left({\rho }^{\prime }V\right)a=Vg(\rho -{\rho }^{\prime })$
$\Rightarrow a=g\left(\frac{\rho -{\rho }^{\prime }}{{\rho }^{\prime }}\right)$ in upward direction $\Rightarrow$ option B is correct
lets say it sinks to the depth $H$, at this instant final velocity becomes $v=0$, so we have
$0=(\sqrt{2gh}{)}^2-2\times g\left(\frac{\rho -{\rho }^{\prime }}{{\rho }^{\prime }}\right)\times H$ $\because v^2=u^2+2as$
$\Rightarrow H=\left(\frac{h{\rho }^{\prime }}{\rho -{\rho }^{\prime }}\right)$ $\Rightarrow$ option C is correct
The body comes back to the surface of lake because when the object is in the liquid, there is a net upward force acting on it. $\Rightarrow$ option D is incorrect