Question

A brass cube of side $a$ and density $\sigma$ is floating in mercury of density $\rho$. If the cube is displaced a bit vertically, it executes S.H.M. 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

Answer: (A)

As a is the side of cube $\sigma$ is its density.
Mass of cube is $a^2\sigma$ its weight $=a^3\sigma g$
Let $h$ be the height of cube immersed in liquid of density $\rho$ in equilibrium then, $F=a^{2}h\rho g=Mg=a^3\sigma g$
If it is pushed down by $y$ then the buoyant force $F^{\prime }=a^2(h+y)\rho g$
Restoring force is $\Delta F=F^{\prime }-F=a^2(h+y)\sigma g-a^{2}h\sigma g$
$=a^{2}y\rho g$
Restoring acceleration $=\frac{\Delta F}{M}=-\frac{a^{2}y\rho g}{M}=-\frac{a^2\rho g}{a^2\sigma }y$
Motion is S.H.M.
$\Rightarrow T=2\pi \sqrt{\frac{a^3\sigma }{a^2\rho g}}=2\pi \sqrt{\frac{a\sigma }{\rho g}}$