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JEE Main 20209 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
9 January
Shift
Morning Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2020 · 9 January · Morning Shift · Physics · Mechanical Properties of Fluids

Water flows m a horizontal tube (see figure). The pressure of water...

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} )

Water flows m a horizontal tube (see figure). The pressure of water changes by 700N m ^ - 2 between A and B where the

Options

  1. A3020cm3/s3020c{m}^{3} / s
  2. B 2720cm3/s2720c{m}^{3} / s
  3. C 2420cm3/s2420c{m}^{3} / s
  4. D 1810cm3/s1810c{m}^{3} / s

Correct answer

B. 2720cm3/s2720c{m}^{3} / s

Step-by-step solution

Using equation of continuity
40VA=20VB40{V}_{A} = 20{V}_{B}
⇒2VA=VB⇒ 2{V}_{A} = {V}_{B}
Using Bernoulli’s equation
PA+12ρVA2=PB+12ρVB2{P}_{A} + \dfrac{1}{2}ρ{V}_{A}^{2} = {P}_{B} + \dfrac{1}{2}ρ{V}_{B}^{2}
⇒PA−PB=12(VB2−VA2)⇒ {P}_{A} - {P}_{B} = \dfrac{1}{2}\left({V}_{B}^{2} - {V}_{A}^{2}\right)
⇒\DeltaP=121000(VB2−VB24)⇒ \DeltaP = \dfrac{1}{2}1000\left({V}_{B}^{2} - \dfrac{{V}_{B}^{2}}{4}\right)
⇒\DeltaP=500×3VB24⇒ \DeltaP = 500 \times \dfrac{3{V}_{B}^{2}}{4}
⇒VB=(\DeltaP)×41500=(700)×41500ms⇒ {V}_{B} = \sqrt{\dfrac{\left(\DeltaP\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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