A wooden cube of side length L floats in water with its four faces vertical. Submerged length of the cube is $l$. When the cube is slightly pushed inside water and released, it under goes oscillation. Then time period of oscillation is 

When the cube is floating $mg=\rho \left(L^{2}l\right)g$.

Where $\rho$ is density of water.

During the course of oscillation if submerged length of the cube is $\left(l+x\right)$, then vertically upward unbalancing force acting on the cube $=\rho L^2(l+x)g-mg=\rho L^{2}xg$

$\therefore$Acceleration, $\frac{d^{2}x}{d{t}^2}=-\frac{\rho L^{2}xg}{m}=\frac{\rho L^{2}xg}{\rho L^{2}l}=-\left(\frac{g}{l}\right)x$

$\therefore {\omega }^2=\frac{g}{l}\Rightarrow T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{l}{g}}$