First slide

Pressure in a fluid-mechanical properties of fluids

Question

A wooden ball of density $ρ$ is immersed in water of density $ρ_{0}$ to depth h and then released. The height H above the surface of water upto which the ball jump out of water is

Moderate

A

zero
B

h
C

$\frac{ρ_{0} h}{ρ}$
D

$(\frac{ρ_{0}}{ρ} - 1) h$

Solution

Let V be the volume of wooden ball. The mass of ball is m = V $ρ$ .
Upward acceleration,
$\begin{array}{l} a = \frac{upward thrust-weight of ball}{mass of ball} \\ \Rightarrow a = \frac{{Vρ}_{0} g - Vρg}{Vρ} = \frac{(ρ_{0} - ρ) g}{ρ} \end{array}$
If v' is the velocity of ball on reaching the surface after being released at depth his
$v = \sqrt{2 as} = {[2 (\frac{p_{0} - ρ}{ρ}) gh]}^{1 / 2}$
If H is the vertical distance reached by ball above the surface of water, then
$\begin{matrix} H = \frac{v^{2}}{2 g} = \frac{2 (ρ_{0} - ρ)}{ρ} gh \times \frac{1}{2 g} \\ = (\frac{ρ_{0} - ρ}{ρ}) h = (\frac{ρ_{0}}{ρ} - 1) h \end{matrix}$

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