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, $F=a^{2} h \rho g=M g=a^{3} \sigma g$
If it is pushed down by $y$ then the buoyant force $F'=a^{2}(h+y) \rho g$
Restoring force is $\Delta F=F'-F=a^{2}(h+y) \sigma g-a^{2} h \sigma g$
$=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.
$\Rightarrow T=2 \pi \sqrt{\frac{a^{3} \sigma}{a^{2} \rho g}}=2 \pi \sqrt{\frac{a \sigma}{\rho g}}$