Question

A container of large uniform cross-sectional area $A$ resting on a horizontal surface, holds two immiscible, non-viscous and incompressible liquids of densities $d$ and $2 d,$ each of height $\frac{H}{2}$ as shown in figure. The lower density liquid is open to the atmosphere having, pressure $P_{0^{'}}$

(i) A homogeneous solid cylinder of length $L (L < \frac{H}{2}),$ crosssectional area $\frac{A}{5}$ is immersed such that if floats with its axis vertical at the liquid-liquid interface with length $\frac{L}{4}$ in the denser liquid. Determine:
(a) the density $D$ of the solid,

(b) the total pressure at the bottom of the container.

(ii) The cylinder is removed and the original arrangement is restored. A tiny hole of area s $(s < < A)$ is, punched on the vertical side of the container at a height $h (h < \frac{H}{2}) .$ Determine:
(a) the initial speed of efflux of the liquid at the hole

(b) the horizontal distance $x$ travelled by the liquid initially, and

(c) the height $h_{m}$ at which the hole should be punched so that the liquid travels the maximum distance) $x_{m}$ initially. Also calculate $x_{m} .$ (Neglect the air resistance in these calculations)

Accepted Answer

(i) (a) Let $ρ$ be the density of cylinder.

As for floating, $Weight = Buoyant Force$

$V ρ g = V_{1} d_{1} g + V_{2} d_{2} g$

or $L (\frac{A}{5}) ρ = (\frac{3}{4} L) (\frac{A}{5}) d + (\frac{1}{4} L) (\frac{A}{5}) 2 d$

i.e., $ρ = \frac{3}{4} d + \frac{2}{4} d = \frac{5}{4} d$

(b) Total pressure $= P_{0} +$ $P_{dense liquid} + P_{light liquid} + P_{buoyancy}$

$P = P_{0} + \frac{1}{A} (A \frac{H}{2} d g + A \frac{H}{2} 2 d g) + \frac{5}{4} d \times (\frac{A}{5} \times L) \times g \times \frac{1}{A}$

i.e. $P = P_{0} + \frac{3}{2} H d g + \frac{1}{4} L d g = P_{0} + \frac{1}{4} (6 H + L) d g$

(ii) (a) By Bernoulli theorem for a point just inside and outside the hole

$P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g h_{1} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$

i.e., $P_{0} + \frac{H}{2} d g + (\frac{H}{2} - h) 2 d g = P_{0} + \frac{1}{2} (2 d) v^{2}$

or $g (3 H - 4 h) = 2 v^{2}$

Thus, velocity of efflux is $v = \sqrt{\frac{g}{2} (3 H - 4 h)}$ .

(b) As at the hole vertical velocity of liquid is zero so time taken by it to reach the ground, $t = \sqrt{(\frac{2 h}{g})}$

So that the horizontal distance covered in time $t$ is given by

$x = v t = \sqrt{\frac{g}{2} (3 H - 4 h)} \times \sqrt{\frac{2 h}{g}} = \sqrt{h (3 H - 4 h)}$

(c) For $x$ to be maximum $x^{2}$ must be maximum,

i.e., $\frac{d}{d h} (x^{2}) = 0$ or $\frac{d}{d h} (3 H h - 4 h^{2}) = 0$

and $d x_{\max} = \sqrt{\frac{3 H}{8} (3 H - \frac{3}{2} H)} = \frac{3}{4} H$

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