A cylindrical tank of cross-sectional area $A = 0.5 m^{2}$ is filled with two liquids of densities $ρ_{1} = 900 kg / m^{3} and ρ_{2} = 600 kg / m^{3}$ to a height h = 60 cm each as shown in the figure. A small hole of area $a = 5 {cm}^{2}$ is made at a height y = 20cm from the bottom in the right vertical wall. Calculate the efflux velocity in m/s. $[Take g = 10 {ms}^{- 2}]$

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Detailed Solution

Let height h of liquid column of density $ρ_{2}$ is equivalent to a height x of liquid column of density $ρ_{1}$

$∴ ρ_{2} gh = ρ_{1} gx; x = \frac{ρ_{2}}{ρ_{1}} \cdot h = \frac{600}{900} \times 0.6 = 0.4 m$

$∴$ The depth of orifice from surface of liquid can be supposed to be,

$h^{'} = [(h - 0.2) + 0.4] = 0.6 - 0.2 + 0.4 = 0.8 m$

$∴$ velocity of efflux, $v = \sqrt{2 gh}$

$= \sqrt{2 \times 10 \times 0.8} = \sqrt{16} = 4 m / s$

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