# A small ball of density ρ is immersed in a liquid of density σ(>ρ) to a depth h and released. The height above the surface of water up to which the ball will jump is

Q: A small ball of density ρ is immersed in a liquid of density σ(>ρ) to a depth h and released. The height above the surface of water up to which the ball will jump is

(a) $\displaystyle (\frac{\sigma}{\rho}-1)h$

(b) $\displaystyle (\frac{\rho}{\sigma}-1)h$

(c) $\displaystyle (\frac{\rho}{\sigma}+1)h$

(c) $\displaystyle (\frac{\sigma}{\rho}+1)h$

Ans: (a)

Sol: Upward Acceleration of the ball in a liquid will be

$\displaystyle a = \frac{F_B – W}{m}$

$\displaystyle a = \frac{V \sigma g – V \rho g}{V \rho}$

$\displaystyle a = (\frac{\sigma}{\rho} – 1) g$

Velocity near the surface of water :

$\displaystyle v = \sqrt{2 a h}$

$\displaystyle v = \sqrt{2 (\frac{\sigma}{\rho} – 1) g h}$

For height above the surface of water :

Let H = height above the surface of water up to which the ball will jump

$\displaystyle H = \frac{v^2 }{2 g }$

$\displaystyle H = \frac{2 (\frac{\sigma}{\rho} – 1) g h }{2 g }$

$\displaystyle = (\frac{\sigma}{\rho}-1)h$