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

By the principle of conservation of energy, we have

$\begin{array}{l} \mathrm{mgh}=\left(\frac{1}{2}\right){\mathrm{mv}}^2\\ \mathrm{or} \mathrm{v}=\sqrt{2\mathrm{gh}}\end{array}$

Since $\mathrm{\rho }>\mathrm{\rho }\text{'}$, the buoyant force of the lake is greater than the weight of the body. If a is the upward acceleration, then

         $\mathrm{ma}=\mathrm{\rho gV}-\mathrm{\rho }\text{'}\mathrm{gV}\text{ or }\left(\mathrm{\rho }\text{'}\mathrm{V}\right)\mathrm{a}=\mathrm{\rho gV}-\mathrm{\rho }\text{'}\mathrm{gV}$

Cancelling V on both the sides, we get

         $\mathrm{a}=\mathrm{g}\left(\frac{\mathrm{\rho }-\mathrm{\rho }\text{'}}{\mathrm{\rho }}\right)=\mathrm{g}\left(\frac{\mathrm{\rho }}{\mathrm{\rho }\text{'}}-1\right)$

Now, ${\mathrm{v}}^2={\mathrm{v}}_0^2-2\mathrm{ay}$

where v₀ is the velocity at y = 0. v = 0 is the velocity at the maximum depth and y is the displacement from the surface to the maximum depth. With these, we get ${\mathrm{v}}_0^2=2\mathrm{ay}$

Hence, the depth is given by

            $\mathrm{y}=\frac{{\mathrm{v}}_0^2}{2\mathrm{a}}=\frac{2\mathrm{gh}}{2\mathrm{g}\frac{\left(\mathrm{\rho }-\mathrm{\rho }\text{'}\right)}{\mathrm{\rho }\text{'}}}=\frac{\mathrm{h\rho }\text{'}}{\mathrm{\rho }-\mathrm{\rho }\text{'}}$