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 $