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(C) The rate of change of the water level $y$ is given by the differential equation: $\frac{dy}{dt} = -k \sqrt{y}$.Separating the variables,we get: $\

Vedclass · Accepted Answer

$60$

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(C) The rate of change of the water level $y$ is given by the differential equation: $\frac{dy}{dt} = -k \sqrt{y}$.Separating the variables,we get: $\frac{dy}{\sqrt{y}} = -k \, dt$.Integrating both sides from the initial depth $y = 4$ to the final depth $y = 0$ over the time interval from $t = 0$ to $t = T$:$\int_{4}^{0} y^{-1/2} \, dy = \int_{0}^{T} -k \, dt$.Evaluating the integrals:$[2\sqrt{y}]_{4}^{0} = -k[t]_{0}^{T}$.Substituting the limits:$2(\sqrt{0} - \sqrt{4}) = -k(T - 0)$.$2(0 - 2) = -kT$.$-4 = -kT$.$T = \frac{4}{k}$.Given $k = \frac{1}{15}$,we have:$T = \frac{4}{1/15} = 4 	imes 15 = 60 	ext{ min}$.

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