Question

A brass cube of side $a$ and density $\sigma$ is floating in mercury of density $\rho$. If the cube is slightly displaced vertically, it executes $SHM$. Its time period will be

• A. $2\pi \sqrt{\frac{\sigma a}{\rho g}}$
• B. $2\pi \sqrt{\frac{\rho a}{\sigma g}}$
• C. $2\pi \sqrt{\frac{\rho g}{\sigma a}}$
• D. $2\pi \sqrt{\frac{\sigma g}{\rho a}}$

Answer 1

The correct option is A $2\pi \sqrt{\frac{\sigma a}{\rho g}}$

As $a$ is the side of the cube and $\sigma$ is its density, the mass of the cube will be
$M=a^3\sigma$
Let $h$ be the height of the cube immersed in the liquid of density $\rho$ in equilibrium.
Then, buoyant force $F=a^{2}h\rho g$
If it is pushed down by depth $y$, then the new buoyant force will be,
$F^{\prime }=a^2(h+y)\rho g$
So, restoring force is $\Delta F=F^{\prime }-F=a^2(h+y)\rho g-a^{2}h\rho g$
$=a^{2}y\rho g$
Restoring acceleration, $a=\frac{-\Delta F}{M}=\frac{-a^{2}y\rho g}{M}=\frac{-a^{2}y\rho g}{a^3\sigma }$
$\Rightarrow a=\frac{-\rho g}{a\sigma }y$
From $a=-{\omega }^{2}y$, w get $\omega =\sqrt{\frac{\rho g}{a\sigma }}$
Thus, the liquid executes SHM.

Time period of oscillation of liquid
$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{a\sigma }{\rho g}}$