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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 liquid of density $ρ$ and mass $m_{0}$ . Assuming that the liquid starts flowing through the hole, the acceleration of the container will be

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

$\frac{a g}{A}$

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

$\frac{A g}{a}$

answer is A.

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Detailed Solution

$F = a ρ v^{2}$

$\begin{array}{l} m_{\circ} a^{1} = a ρ (2 g h) \\ a^{1} = \frac{(a ρ 2 g h)}{(A h ρ)} = \frac{2 a g}{A} \end{array}$

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