Question

The density $\rho$ of a liquid varies with depth $h$ from the free surface as $\rho =kh$. A small body of density ${\rho }_1$ is released from the surface of the liquid. The body executes $SHM$ with amplitude

• A. $\frac{2{\rho }_1}{k}$
• B. $\frac{{\rho }_1}{k}$
• C. $\frac{\sqrt{2}{\rho }_1}{k}$
• D. $\frac{\sqrt{3}{\rho }_1}{k}$

Answer 1

The correct option is B $\frac{{\rho }_1}{k}$

Let V be the volume of body which is released on the surface of liquid. When body reaches at depth $h$ from the surface of liquid, then upward thrust on body due to liquid $=V\left(kh\right)g$

Weight of body $=V{\rho }_{1}g$

At equilibrium, `V(kx)g=Vrho_(1)g or x=rho_(1)//k`