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Q.

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 p and mass mo. Assuming that the liquid starts flowing though the hole, the acceleration of the container will be

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a

$\frac{2 Ag}{a}$

b

$\frac{ag}{A}$

c

$\frac{2 ag}{A}$

d

$\frac{ag}{A^{2}}$

answer is A.

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Detailed Solution

see fig (11)

Question Image

Here $m_{0} f = \frac{dp}{dt}$
$or (AHP) f = (Vρ) V$
where f = acceleration, V = volume of liquid flowing per second and v = velocity of flow.
Since V = $α$ v, we have
$f = \frac{(avρ) v}{AHρ} = \frac{{av}^{2}}{AH} or f = \frac{a (2 gH)}{AH} = \frac{2 ag}{A}$

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