First slide

Fluid dynamics

Question

A tank is filled with water up to a height H. Water is allowed to come out of a hole P in one of the walls at a depth h below the surface of water . Express the horizontal distance x in terms of H and h

Moderate

A

$x = \sqrt{h (H - h)}$
B

$x = \sqrt{[h (H - h) / 2]}$
C

$x = 2 \sqrt{h (H - h)}$
D

$X = 4 \sqrt{h (H - h)}$

Solution

The pressure at the free surface of the liquid and also at outside of point P is atmospheric. Hence there will be no effect of atmospheric pressure on the flow of liquid from hole P. The liquid on the free surface has no kinetic energy but only potential energy. On the other hand, the Liquid coming out of the hole has both kinetic and potential energies. Let v be the velocity of efflux of the liquid coming out from the hole. According to Bernoulli's theorem, we having
$P + 0 + ρgH = P + ρg (H - h) + \frac{1}{2} {ρv}^{2} V = \sqrt{(2 gh)} After coming ftom the hole, the ltquid adopts a parabolic path . II it takes t \sec in falling through a vertical distance [] I - i), then (H - h) = \frac{1}{2} {gt}^{2} (∵ s = ut + \frac{1}{2} {gt}^{2}) ∴ t = \sqrt{{2 (H - h) / g}} From eqs. (1) and (2), X = vt = 2 \sqrt{{h (H - h)}}$

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