First slide

Fluid statics

Question

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

Moderate

A

$(\frac{σ}{ρ} - 1) h$
B

$(\frac{ρ}{σ} - 1) h$
C

$(\frac{ρ}{σ} + 1) h$
D

$(\frac{σ}{ρ} + 1) h$

Solution

Let V be the volume of the ball.
Net upward force = $Vσg - Vρg$
Net upward acceleration,
$a = \frac{Vσg - Vρg}{Vρ} = \frac{(σ - ρ) g}{ρ}$
Velocity at the surface = $\sqrt{\frac{2 (σ - ρ) gh}{ρ}}$
If $h_{a}$ is the height in air to which the ball rises, then
$0 - \frac{2 (σ - ρ) gh}{ρ} = 2 (- g) h_{a}$
$h_{a} = \frac{(σ - r) gh}{gρ} = (\frac{σ}{ρ} - 1) h$

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