Question

A long metal rod of length $l$ and 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

• A. $\sqrt{2gl}$
• B. $\sqrt{2gl\sigma }$
• C. $\sqrt{2gl(1-\frac{1}{2\sigma })}$
• D. $\sqrt{2gl(2\sigma -1)}$

Answer 1

The correct option is C

$\sqrt{2gl(1-\frac{1}{2\sigma })}$

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.
$\frac{dv}{dt}=\frac{\pi r^{2}l\rho g-\pi r^{2}x{\rho }_{0}g}{\pi r^{2}l\rho }=g-\frac{gx{\rho }_0}{l\rho }=g(1-\frac{x}{\sigma l})$
$\Rightarrow \frac{dv}{dx}.\frac{dx}{dt}=g(1-\frac{x}{\sigma l})$
$\Rightarrow v\frac{dv}{dx}=g(1-\frac{x}{\sigma l})$
Integrating, we have
$\frac{v^2}{2}=g{|x-\frac{x^2}{2\sigma l}|}_0^l=gl(1-\frac{1}{2\sigma })$
$\Rightarrow v=\sqrt{2gl(1-\frac{1}{2\sigma })}$
Hence, the correct choice is (c).