First slide

Pressure in a fluid-mechanical properties of fluids

Question

A solid uniform ball of volume V floats on the interface of two immiscible liquids (see the figure). The specific gravity of the upper liquid $(H_{2} O)$ is $ρ_{1}$ and that of lower one $(H g)$ is $ρ_{2}$ and the specific gravity of ball is $ρ (ρ_{1} < ρ < ρ_{2})$ . The fraction of the volume of the ball in the upper liquid is

Moderate

A

$\frac{ρ_{2}}{ρ_{1}}$
B

$\frac{ρ_{2} - ρ}{ρ_{2} - ρ_{1}}$
C

$\frac{ρ - ρ_{1}}{ρ_{2} - ρ_{1}}$
D

$\frac{ρ_{1}}{ρ_{2}}$

Solution

Let V be the total volume of the ball and v be the volume of the ball in the upper liquid. Then (V - v) is the volume of the lower liquid displaced.
Using the law of floatation, we have
$\begin{array}{l} m g = B \\ \Rightarrow V ρ g = v ρ_{1} g + (V - v) ρ_{2} g \\ \Rightarrow V ρ = v ρ_{1} + V ρ_{2} - v ρ_{2} \\ ⇒ V (ρ - ρ_{2}) = v (ρ_{1} - ρ_{2}) \\ \Rightarrow \frac{v}{V} = \frac{ρ - ρ_{2}}{ρ_{1} - ρ_{2}} = \frac{ρ_{2} - ρ}{ρ_{2} - ρ_{1}} \end{array}$

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