Question

A wooden cube (density of wood 'd') of side '$\ell$' floats in a liquid '$\rho$' with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, it performs simple harmonic motion of period 'T'. Then, 'T' is equal to :-

• A. $2\pi =\sqrt{\frac{\ell \rho }{(\rho -d)g}}$
• B. $2\pi =\sqrt{\frac{\ell d}{\rho g}}$
• C. $2\pi =\sqrt{\frac{\ell \rho }{\rho g}}$
• D. $2\pi =\sqrt{\frac{\ell d}{(\rho -d)g}}$

Answer 1

The correct option is B $2\pi =\sqrt{\frac{\ell d}{\rho g}}$

Let at any instant, the cube be at a depth x from the equilibrium position,

then Net force acting on the cube = Upthrust on the portion of length x .

Therefore $F=-\rho l^{2}xg$-------(i)

where $F$ is the force exerted , $\rho$ is the density of the liquid $l^2\times x$ is the volume and $g$ is the acceleration due to gravity .

It performs simple harmonic motion and it can be seen that ,

$F\propto -x$

As the sign is negative it is implied that the force is applied in the opposite direction .

Therefore the equation for simple harmonic motion is

$F=-kx$ where $k$ is a constant . --------(ii)

Comparing the value of $k$ with (i)

We get , $k=\rho l^{2}g$

Now time period $T$ is equal to

$T=2\pi \sqrt{\frac{m}{k}}$

Or, $T=2\pi \sqrt{\frac{l^{3}d}{\rho l^{2}g}}$ \\ where $d$ is the density of the cube

Or,$T=2\pi \sqrt{\frac{ld}{\rho g}}$ (ANSWER)