A small hole of area a is at the bottom of a container of area A. The liquid is filled up to height h from base. As liquid comes out then $[a = \frac{A}{3}]$ .

Question

A small hole of area a is at the bottom of a container of area A. The liquid is filled up to height h from base. As liquid comes out then [ a = 3 A ​ ] .

TestHub · Accepted Answer

Applying Bernoulli's equation between P (top surface) and Q (hole):

$p_{0} + ρ g h + \frac{1}{2} ρ v_{1}^{2} = p_{0} + 0 + \frac{1}{2} ρ v_{2}^{2} \Rightarrow v_{2}^{2} - v_{1}^{2} = 2 g h$

From equation of continuity, $A v_{1} = a v_{2} = \frac{A}{3} v_{2} \Rightarrow v_{2} = 3 v_{1}$

$9 v_{1}^{2} - v_{1}^{2} = 2 g h \Rightarrow v_{1} = \frac{g h}{4} = \frac{g h}{2}$ so (A) is correct.

Acceleration of the top layer: $a = \frac{d v _{1}}{d t} = \frac{1}{2} g \times \frac{1}{2 h} \times \frac{d h}{d t} = - \frac{1}{4} \frac{g}{h} \times \frac{g h}{2} = - \frac{g}{8} m / s^{2}$

Magnitude of acceleration is g/8 m/s², so (B) is correct and (C) is wrong.

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