Fluid Mechanics

JEE Main level, single correct, medium

65 of 192

Question 65 of 192

Water flows m a horizontal tube (see figure). The pressure of water changes by 700Nm−2700N{m}^{-2} between AA and BB where the area of cross section are 40cm240c{m}^{2} and 20cm2,20c{m}^{2}, respectively. Find the rate of flow of water through the tube. (density of water =1000kgm−3=1000kg{m}^{-3} )

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Answer: (B) 2720cm3/s2720c{m}^{3}/s

Using equation of continuity
40VA=20VB40{V}_{A}=20{V}_{B}
⇒2VA=VB\Rightarrow 2{V}_{A}={V}_{B}
Using Bernoulli’s equation
PA+12ρVA2=PB+12ρVB2{P}_{A}+\dfrac{1}{2}\rho {V}_{A}^{2}={P}_{B}+\dfrac{1}{2}\rho {V}_{B}^{2}
⇒PA−PB=12(VB2−VA2)\Rightarrow {P}_{A}-{P}_{B}=\dfrac{1}{2}\left({V}_{B}^{2}-{V}_{A}^{2}\right)
⇒ΔP=121000(VB2−VB24)\Rightarrow \Delta P=\dfrac{1}{2}1000\left({V}_{B}^{2}-\dfrac{{V}_{B}^{2}}{4}\right)
⇒ΔP=500×3VB24\Rightarrow \Delta P=500\times \dfrac{3{V}_{B}^{2}}{4}
⇒VB=(ΔP)×41500=(700)×41500ms\Rightarrow {V}_{B}=\sqrt{\dfrac{\left(\Delta P\right)\times 4}{1500}}=\sqrt{\dfrac{\left(700\right)\times 4}{1500}}\dfrac{m}{s}
Volume flow rate =20×100×VB=2732cm3/s=20\times 100\times {V}_{B}=2732c{m}^{3}/s

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