A long metal rod of length land relative density $\sigma$ is held vertically with its lower end just touching the surface of water. The speed of the rod when it just sinks in water is given by

Let the densities of metal and water be $\rho$ and ${\rho }_0$ respectively and let x be the length of the rod immersed in water at an instant of time t. Then, acceleration at that instant = apparent weight divided by mass of the rod, i.e. 

        $$\begin{gathered} \frac{dv}{dt}=\frac{{\mathrm{\pi r}}^2\mathrm{l\rho g}-{\mathrm{\pi r}}^2{\mathrm{x\rho }}_0\mathrm{g}}{{\mathrm{\pi r}}^2\mathrm{l\rho }}=g-\frac{gx{\rho }_0}{l\rho } \\ =g\left(1-\frac{x}{\sigma l}\right) \end{gathered}$$

or  $\frac{dv}{dx}·\frac{dx}{dt}=g\left(1-\frac{x}{\sigma l}\right)$

or   $v\frac{dv}{dx}=g\left(1-\frac{x}{\sigma l}\right)$

Integrating, we have

        $\frac{v^2}{2}=g{{\left|x-\frac{x^2}{2\sigma l}\right|}_0}^l$=$g\left(l-\frac{l}{2\sigma }\right)$

or    $v=\sqrt{2gl\left(1-\frac{1}{2\sigma }\right)}$which is choice(3).