There is a small hole in the bottom of a fixed container containing a liquid up to a height h. The top of the liquid as well as the hole at the bottom are exposed to atmosphere. As the liquid comes out of the hole, (area of the hole is a and that of the top surface is A) Assume a < < A

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The top surface of the liquid accelerates with acceleration $= g \frac{a^{2}}{A^{2}}$

The top surface of the liquid accelerates with acceleration = g

The top surface of the liquid retards with retardation $= g \frac{a}{A}$

The top surface of the liquid retards with retardation $= \frac{g a^{2}}{A^{2}}$

answer is D.

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Detailed Solution

The velocity of fluid at the hole is

$v_{2} = \sqrt{\frac{2 g h}{1 + (a^{2} / A^{2})}}$

Using continuity equation at the two cross-sections (1) and (2) :

$v_{1} A = v_{2} a \Rightarrow v_{1} = \frac{a}{A} v_{2}$

$\Rightarrow acceleration (of top surface) =_{- v_{1}} \frac{d v_{1}}{d h}$

$= - \frac{a}{A} v_{2} \frac{d}{d h} (\frac{a}{A} v_{2})$

$a_{1} = - \frac{a^{2}}{A^{2}} v_{2} \frac{d v_{2}}{d h} = - \frac{a^{2}}{A^{2}} \frac{\sqrt{2 g h} \sqrt{2 g}}{(1 + \frac{a^{2}}{A^{2}})} \cdot \frac{1}{2 \sqrt{h}} (∵ A ≫> a)$

$(1 >> \frac{a^{2}}{A^{2}}) .$ so neglecting $\frac{a^{2}}{A^{2}}$ we get

$\Rightarrow a_{1} = \frac{- g a^{2}}{A^{2}}$

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