A rubber ball of mass m and density ρ is immersed in a liquid of density 3ρ to a depth h and released…

Q: A rubber ball of mass m and density ρ is immersed in a liquid of density 3ρ to a depth h and released. To what height will the ball jump up above the surface due to buoyancy force of liquid on the ball ? (neglect the resistance of water and air).

Sol: Volume of the ball $\large V = \frac{m}{\rho}$

Acceleration of ball moving in upward direction inside the liquid

$\large a = \frac{F_{net}}{m} $

$\large a = \frac{upthrust – weight}{m} = \frac{V \rho_l g – m g}{m}$

$\large a = \frac{\frac{m}{\rho} \times 3 \rho g – m g}{m}$

a = 2 g (upwards)

velocity of ball while crossing the surface

$\large v = \sqrt{2 a h} = \sqrt{4 g h}$

The ball will jump to a height

$\large H = \frac{v^2}{2 g} = \frac{4 g h}{2 g}$

H = 2 h