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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_o. Assuming that the liquid starts flowing through the small hole, the acceleration of the container will be

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$\frac{2 ag}{A}$

$\frac{ag}{A}$

$\frac{2 Ag}{a}$

$\frac{Ag}{a}$

answer is A.

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Detailed Solution

$\begin{array}{l} F = {Aρu}^{2}; {ma}^{1} = {aρu}^{2} \\ a^{1} = \frac{{aρu}^{2}}{m} = \frac{aρ \times 2 gh}{m} \\ ∴ a^{1} = \frac{aρ 2 gh}{Ahρ} = \frac{2 ag}{A} \end{array}$

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