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A cylindrical tank of base area A has a small hole of area a the bottom. At time t=0, a tap starts to supply water into the tank at a constant rate ${αm}^{3} / s$ .
find the time when level of water becomes $h (< h_{\max})$ .

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$\frac{A}{ag} [\frac{α}{a} \ln \{\frac{α - a \sqrt{9 gh}}{α}\} - \sqrt{5 gh}]$

$\frac{A}{ag} [\frac{α}{a} \ln \{\frac{α - a \sqrt{12 gh}}{α}\} - \sqrt{6 gh}]$

$\frac{A}{ag} [\frac{α}{a} \ln \{\frac{α - a \sqrt{2 gh}}{α}\} - \sqrt{2 gh}]$

answer is A.

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Detailed Solution

Let at time t, the level of water be h. Then,

$A (\frac{dh}{dt}) = α - a \sqrt{2 gh} or \int_{0}^{h} \frac{dh}{α - a \sqrt{2 gh}} = \int_{0}^{t} \frac{dt}{A} Solving this, we get t = \frac{A}{ag} [\frac{α}{a} \ln \{\frac{α - a \sqrt{2 gh}}{α}\} - \sqrt{2 gh}]$

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