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A large open top container of negligible mass and uniform cross-sectional area A has a small hole of cross sectional area ‘a’ in its side wall near the bottom. The container is kept over a smooth horizontal floor and contains a liquid of density $ρ$ and mass $m_{0}$ assuming that the liquid starts flowing through the hole a. The acceleration of the container will be

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$\frac{A g}{a}$

$\frac{2 A g}{a}$

$\frac{2 a g}{A}$

$\frac{a g}{A}$

answer is A.

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Detailed Solution

$F = A ρ v^{2} \Rightarrow m a^{1} = a ρ v^{2} \Rightarrow a^{1} = \frac{a ρ v^{2}}{m} \Rightarrow a^{1} = \frac{a ρ 2 g h}{m}$

$∴ a^{1} = \frac{a ρ 2 g h}{A h ρ} = \frac{2 a g}{A}$

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