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 oscillation with a time period T. Then :

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

Answer 1

Answer: (B)

T $\propto$ $\bigg(\frac{1}{\sqrt A}\bigg)$

Answer 2

Time period of SHM is given by
T= $2\pi\frac{\sqrt I}{g}$
according to law of floatation
weight of the block = weight of the liquid displaced

mg=Al\rho$$g

l= $\frac{m}{A\rho}$

Then, T=$2\pi\frac{\sqrt m}{A\rho g}$

Now we can say that-

T $\propto$ $\bigg(\frac{1}{\sqrt A}\bigg)$

Therefore, the correct option is (B): T $\propto$ $\bigg(\frac{1}{\sqrt A}\bigg)$