A small wooden ball of density ρ is immersed in water of density sigma to depth h and then released

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9-1.Fluid Mechanics

normal

A small wooden ball of density $ \rho$ is immersed in water of density $\sigma $ to depth $h $ and then released. The height $H$ above the surface of water up to which the ball will jump out of water is

A

$\frac{{\sigma h}}{\rho }$

B

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

C

$h$

D

zero

Solution

Net upward force acting on the body $F=\sigma V g-\rho V g$

$\Rightarrow m a=\sigma V g-\rho V g$

$\Rightarrow(\rho V) a=V g(\sigma-\rho)$

$\Rightarrow a=g\left(\frac{\sigma}{\rho}-1\right)$

Velocity $v$ after height $h=\sqrt{2 g\left(\frac{\sigma}{\rho}-1\right) h}$

Conserving mechanical energy at max height h'above surface,

$m g h^{\prime}=\frac{1}{2} m\left(2 g\left(\frac{\sigma}{\rho}-1\right) h\right)$

$\Rightarrow h^{\prime}=\left(\frac{\sigma}{\rho}-1\right) h$

Standard 11

Physics

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