Question

a tank of cross section A contains a liquid of density up to a height H.there is a hole of cross section area A,at the bottom of the tank.the time in which the liquid level us reduced to half is?​

Answer 1

Answer:

The time in which the liquid level us reduced to half is

$t = \frac{2A}{a}\sqrt{\frac{H}{2g}}(1 - \frac{1}{\sqrt2})$

Explanation:

As we know by the equation of continuity

$A_1v_1 = A_2v_2$

now we have

$A(-\frac{dy}{dt}) = a\sqrt{2gy}$

now we have

$-A\int \frac{dy}{\sqrt y} = a\sqrt{2g} \int dt$

now we have

$- A(2\sqrt y) = a\sqrt{2g} t$

now level changes from H to H/2 so we have

$2A(\sqrt{H} - \sqrt{\frac{H}{2}}) = a\sqrt{2g} t$

now we have

$t = \frac{2A}{a}\sqrt{\frac{H}{2g}}(1 - \frac{1}{\sqrt2})$

#Learn

Topic : Equation of continuity

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