Two large, identical water tanks, 1 and 2, kept on the top of a building of height $H$, are filled with…
Solution
$\begin{aligned} & \mathrm{Av}=\mathrm{av}_1 \\ & \mathrm{~A}\left(-\frac{\mathrm{dy}}{\mathrm{dt}}\right)=\mathrm{a} \sqrt{2 \mathrm{gy}} ; \mathrm{dt}=\frac{\mathrm{A}}{\mathrm{a} \sqrt{2 \mathrm{~g}}} \cdot \frac{-d y}{\sqrt{y}} \\ & \int_0^{t_1} \mathrm{dt}=\frac{\mathrm{A}}{\mathrm{a} \sqrt{2 \mathrm{~g}}} \int_{\mathrm{h}}^0-\frac{d y}{\sqrt{y}} \\ & t_1=\frac{\mathrm{A}}{\mathrm{a} \sqrt{2 \mathrm{~g}}} 2 \sqrt{\mathrm{h}} ; \mathrm{t}_1=\frac{\mathrm{A}}{\mathrm{a}} \sqrt{\frac{2 h}{g}}\end{aligned}$
$A v^{\prime}=\mathrm{av}_2$
$\begin{aligned} & A\left(-\frac{d y}{d t}\right)=a \sqrt{2 g(H+y)} \\ & d t=-\frac{A}{a \sqrt{2 g}} \frac{d y}{\sqrt{H+y}} \\ & \int_0^{t_2} d t=-\frac{A}{a \sqrt{2 g}} \int_H^0 \frac{d y}{\sqrt{H+y}} \\ & t_2=\frac{A}{a \sqrt{2 g}}(2)(\sqrt{H+h}-\sqrt{H}) \quad \& \quad H=\frac{16 h}{9} \\ & =\frac{A}{a} \sqrt{\frac{2 h}{g}}\left(\frac{5}{3}-\frac{4}{3}\right) \\ & t_2=\frac{A}{a} \sqrt{\frac{2 h}{g}}\left(\frac{1}{3}\right)\end{aligned}$
ratio $\frac{\mathrm{t}_1}{\mathrm{t}_2}=3$
^Asked in: JEE Advanced 2024 (Paper 1)
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