First slide

Pressure in a fluid-mechanical properties of fluids

Question

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

Difficult

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 $= V σ g - V ρ g$
Net upward acceleration,
$a = \frac{V σ g - V ρ g}{V ρ} = \frac{(σ - ρ) g}{ρ}$
$Velocity at the surface = \sqrt{\frac{2 (σ - ρ) g h}{ρ}}$
If h_a is the height in air to which the ball rises, then
$\begin{array}{l} 0 - \frac{2 (σ - ρ) g h}{ρ} = 2 (- g) h_{a} \\ ∴ h_{a} = \frac{(σ - r) g h}{g ρ} = (\frac{σ}{ρ} - 1) h \end{array}$

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