A brass cube of side $a$ and density $\rho$ is floating in mercury of density $\sigma$. If the cube is displaced a bit vertically, it executes $S.H.M.$ Its time period will be
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(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 $\sigma$ 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 $F' = {a^2}(h + y)\rho \,g$

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 $S.H.M.$

==> $T = 2\pi \sqrt {\frac{{{a^3}\sigma }}{{{a^2}\rho \,g}}} = 2\pi \sqrt {\frac{{a\sigma }}{{\rho \,g}}} $

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