Physics - Fluid Question with Solution | TestHub

Mass of the cube is $m=l^{3}d$.

When the cube is pushed down by a small distance $x$, the extra volume submerged in the liquid is $\Delta V=l^{2}x$.

The restoring force acting on the cube is the additional buoyant force:

$F=-(\Delta V)\rho g=-l^{2}x\rho g$

Using Newton's second law, $F=ma$:

$l^{3}d\cdot a=-l^2\rho gx$

$a=-\left(\frac{\rho g}{ld}\right)x$

Comparing this with the standard equation of simple harmonic motion $a=-{\omega }^{2}x$, we get:

${\omega }^2=\frac{\rho g}{ld}\Rightarrow \omega =\sqrt{\frac{\rho g}{ld}}$

The time period $T$ is given by:

$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{ld}{\rho g}}$

Answer: $2\pi \sqrt{\frac{ld}{\rho g}}$