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Q.

A container of large uniform cross – sectional area holds two immiscible, non – viscous and incompressible liquids of densities $ρ$ and $4 ρ$ , each of height $\frac{H}{2}$ . Determine the position where, a small hole has to be punched, so that the heavier liquid comes out with a maximum range R initially.

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a

At a height $\frac{H}{4}$ from the bottom

b

At a height $\frac{5}{4} H$ from the bottom

c

At a height $\frac{5}{16} H$ from the bottom

d

At a height $\frac{7}{18} H$ from the bottom

answer is C.

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Detailed Solution

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On applying Bernoulli’s theorem between points (1) and (2), we have

$p_{0} + ρ g \frac{H}{2} + 4 ρ g (\frac{H}{2} - h) = p_{0} + \frac{1}{2} 4 ρ u^{2}$

$ρ g \frac{H}{2} + 2 ρ g H - 4 ρ g h = 2 ρ u^{2}$

$\frac{5 ρ g H}{2} - 4 ρ g h = 2 ρ u^{2}$ $\Rightarrow u^{2} = [\frac{5}{4} g H - 2 g h]$

$\Rightarrow u^{2} = \frac{g}{4} [5 H - 8 h]$ $\Rightarrow u = \frac{\sqrt{g}}{2} \sqrt{5 H - 8 h}$ $t = \sqrt{\frac{2 h}{g}}$

$R = u t = \frac{\sqrt{g}}{2} \sqrt{5 H - 8 h} . \sqrt{\frac{2 h}{g}} = \frac{1}{\sqrt{2}} \sqrt{5 H h - 8 h^{2}}$

For R to be maximum,

$\frac{d R}{d h} = 0$ $\frac{d R}{d h} = \frac{1}{\sqrt{2}} . \frac{1}{2 \sqrt{5 H h - 8 h^{2}}} (5 H - 16 h) = 0$

$\Rightarrow 5 H - 16 h = 0$ $\Rightarrow h = \frac{5}{16} H$

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