Water is drained from a vertical cylindrical | vedclass

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(C) The rate at which the water level $y$ drops is given by the differential equation: $\frac{dy}{dt} = -k\sqrt{y}$.Given $k = \frac{1}{15}$,the equat

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$60$

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(C) The rate at which the water level $y$ drops is given by the differential equation: $\frac{dy}{dt} = -k\sqrt{y}$.Given $k = \frac{1}{15}$,the equation becomes $\frac{dy}{dt} = -\frac{1}{15}\sqrt{y}$.To find the time $t$ required to drain the tank from an initial depth of $4 	ext{ m}$ to $0 	ext{ m}$,we separate the variables and integrate:$\int_{4}^{0} \frac{dy}{\sqrt{y}} = \int_{0}^{t} -\frac{1}{15} dt$.Evaluating the integrals:$[2\sqrt{y}]_{4}^{0} = -\frac{1}{15}t$.Substituting the limits:$2(\sqrt{0} - \sqrt{4}) = -\frac{t}{15}$.$2(0 - 2) = -\frac{t}{15}$.$-4 = -\frac{t}{15}$.$t = 4 	imes 15 = 60 	ext{ minutes}$.

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