Fluid Mechanics

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Question 72 of 192


In the diagram shown, the difference in the two tubes of the manometer is 5cm5cm, the cross-section of the tube at AA and BB is 6mm26{mm}^{2} and 10mm210{mm}^{2} respectively. The rate at which water flows through the tube is (g=10ms−2)\left(g=10m{s}^{-2}\right)

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Answer: (A) 7.5ccs−17.5cc{s}^{-1}

6vA=10vB......(1)6{v}_{A}=10{v}_{B}......\left(1\right)
12ρvA2−ρgh=12ρvB2\dfrac{1}{2}\rho {v}_{A}^{2}-\rho gh=\dfrac{1}{2}\rho {v}_{B}^{2} .........(2)
⇒vA=10.8ms−1\Rightarrow {v}_{A}=\dfrac{1}{0.8}m{s}^{-1}
dVdt=AvA\dfrac{dV}{dt}=A{v}_{A}
=6×10−2cm2×1000.8cms−1=7.5ccs−1=6\times {10}^{-2}{cm}^{2}\times \dfrac{100}{0.8}cm{s}^{-1}=7.5cc{s}^{-1}
OR
By the concept of the venturi meter tube
rate of flow=Q=A1A22g(h1−h2)(A22−A12)=Q={A}_{1}{A}_{2}\sqrt{\dfrac{2g({h}_{1}-{h}_{2})}{({A}_{2}^{2}-{A}_{1}^{2})}}
=6×10×10−42×100×5(−36+100)×10−2=6\times 10\times {10}^{-4}\sqrt{\dfrac{2\times 100\times 5}{(-36+100)\times {10}^{-2}}}
=7.5ccs−1=7.5cc{s}^{-1}

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