Question

A rectangular block of mass $m$ and area of cross-section A floats in a liquid of density $\rho$. If it is given a small vertical displacement from equilibrium it undergoes with a time period $T$. Then

• A. $T\propto \frac{1}{\sqrt{m}}$
• B. $T\propto \sqrt{\rho }$
• C. $T\propto \frac{1}{\sqrt{A}}$
• D. $T\propto \frac{1}{\rho }$

Answer 1

Answer: (C)

$T\propto \frac{1}{\sqrt{A}}$

Refer figure, Let h be the height of block immersed in liquid, when the block is floating.
$mg=Ah\rho g$ ...(i)
If the block is given a vertical displacement y, then the effective restoring force is
$F=-\left[A\right(h+y)\rho g-mg]=-\left[A\right(h+y)\rho g-Ah\rho g]$
=$-A\rho gy$ ....... (ii)
i.e. $F\propto y$ and -ve sign shows that F is directed towards its equilibrium position of block. So, if the block is left free, it will execute SHM
For SHM, $F=-m{\omega }^{2}y$ ...(iii)
Comparing (ii) and (iii), we get
$A\rho g=m{\omega }^2$, $\omega =\sqrt{\frac{A\rho g}{m}}$

$\therefore Timeperiod,T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{A\rho g}}\to T\propto \frac{1}{\sqrt{A}}$