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JEE Advanced 2024Paper 1
Exam
JEE Advanced
Year
2024
Paper
Paper 1
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Advanced 2024 · Paper 1 · Physics · Mechanical Properties of Fluids

Two large, identical water tanks, 1 and 2, kept on the top of a...

Two large, identical water tanks, 1 and 2, kept on the top of a building of height HH, are filled with water up to height hh in each tank. Both the tanks contain an identical hole of small radius on their sides, close to their bottom. A pipe of the same internal radius as that of the hole is connected to tank 2 , and the pipe ends at the ground level. When the water flows from the tanks 1 and 2 through the holes, the times taken to empty the tanks are t1t_1 and t2t_2, respectively. If H=(169)hH=\left(\frac{16}{9}\right) h, then the ratio t1/t2t_1 / t_2 is _______ .Two large, identical water tanks, 1 and 2, kept on the top of a building of height H , are filled with water up toTwo large, identical water tanks, 1 and 2, kept on the top of a building of height H , are filled with water up to

Correct answer

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Step-by-step solution

Two large, identical water tanks, 1 and 2, kept on the top of a building of height H , are filled with water up to Av=av1 A(−dydt)=a2gy;dt=Aa2 g⋅−dyy∫0t1dt=Aa2 g∫h0−dyyt1=Aa2 g2h;t1=Aa2hg\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}Two large, identical water tanks, 1 and 2, kept on the top of a building of height H , are filled with water up to Av′=av2A v^{\prime}=\mathrm{av}_2A(−dydt)=a2g(H+y)dt=−Aa2gdyH+y∫0t2dt=−Aa2g∫H0dyH+yt2=Aa2g(2)(H+h−H)H=16h9=Aa2hg(53−43)t2=Aa2hg(13)\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
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