Question

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

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

Answer 1

The correct option is B $T\propto \frac{1}{\sqrt{A}}$

The upthrust on the block (restoring force on block)

$=-$The force applied on the box

$=-$mass of liquid displaced$\times g$

$=-$Volume of liquid displaced$\times$density of liquid$\times g$

$=-$Surface area block$\times$vertical displacement of block $\times \overline{n}g$

$=-Ax\overline{n}g$ (negative sign signifies that it is in upward direction)

$\Rightarrow ma=-Ax\overline{n}g$, ($m=$Mass of acceleration, $a=$acceleration of block)

$\Rightarrow a=-\frac{Ax\overline{n}g}{m}$

$\Rightarrow a+{\omega }^{2}x=0$ (Here $\omega =\sqrt{\frac{Ax\overline{n}g}{m}}$=Angular frequency)

Time period, $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{A\overline{n}g}}$

So, $T\propto \frac{1}{\sqrt{A}}$ and $T\propto \sqrt{\frac{1}{\overline{n}}}$