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 $