Question

A small ball of density $\rho$ is immersed in a liquid of density $(\sigma > \rho)$ to a depth h and released. The height above the surface of water up to the ball will jump is

• A. $\left(\frac{\rho}{\sigma}-1\right)h$
• B. $\left(\frac{\rho}{\sigma}+1\right)h$
• C. $\left(\frac{\sigma}{\rho}-1\right)h$
• D. $\left(\frac{\sigma}{\rho}+_1\right)h$

Answer 1

Answer: (C)

$\left(\frac{\sigma}{\rho}-1\right)h$

Answer 2

$a = \left( \frac{ \sigma - \rho}{\rho} \right) g$ $v = \sqrt{\frac{2( \sigma - \rho) gh}{\rho}}$ $\therefore H = \frac{v^2}{2g} = \left( \frac{ \sigma - \rho}{\rho} \right) h$