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Water is drained from a vertical cylindrical tank by opening a valve at the base tank. It is known that the rate at which the water level drops is proportional to the square root of water depth y, where the constant of proportionally k > 0 depends on the acceleration due to gravity and the geometry of the hole. If time t is measured in minute and $k = \frac{1}{15}$ then the time to drain the tank if the water is 4 m deep to start with is

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60 min

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80 min

answer is C.

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Detailed Solution

According to the question
$\frac{dy}{dt} = - k \sqrt{y} \begin{array}{l} Or \int_{4}^{0} \frac{dy}{\sqrt{y}} = - k \int_{0}^{t} dt \\ 0 - 4 = - \frac{t}{15} \end{array}$
Or t = 60 min

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