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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Fluid Mechanics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Fluid Mechanics

A tube with cross sections at A and B , having areas of 4 mm^2 and...

A tube with cross sections at AA and BB, having areas of 4 mm24 \text{ mm}^2 and 2 mm22 \text{ mm}^2 respectively, is kept vertical with BB positioned upward. The distance separating AA and BB is 1516 cm\dfrac{15}{16} \text{ cm}. Water enters through BB at a rate of 1 cm3 s−11 \text{ cm}^3 \text{ s}^{-1}. Determine the pressure difference PA−PBP_A - P_B. Note that the speed decreases as the water falls down. Take g=10 m s−2g = 10 \text{ m s}^{-2}.

Options

  1. A188 N/m2188 \text{ N/m}^2
  2. B94 N/m294 \text{ N/m}^2
  3. C282 N/m2282 \text{ N/m}^2
  4. D150 N/m2150 \text{ N/m}^2

Correct answer

A. 188 N/m2188 \text{ N/m}^2

Step-by-step solution

Volume flow rate Q=1 cm3 s−1=10−6 m3 s−1Q = 1 \text{ cm}^3 \text{ s}^{-1} = 10^{-6} \text{ m}^3 \text{ s}^{-1}

Area at AA, AA=4 mm2=4×10−6 m2A_A = 4 \text{ mm}^2 = 4 \times 10^{-6} \text{ m}^2

Area at BB, AB=2 mm2=2×10−6 m2A_B = 2 \text{ mm}^2 = 2 \times 10^{-6} \text{ m}^2

Velocity at AA, vA=QAA=10−64×10−6=0.25 m/sv_A = \dfrac{Q}{A_A} = \dfrac{10^{-6}}{4 \times 10^{-6}} = 0.25 \text{ m/s}

Velocity at BB, vB=QAB=10−62×10−6=0.5 m/sv_B = \dfrac{Q}{A_B} = \dfrac{10^{-6}}{2 \times 10^{-6}} = 0.5 \text{ m/s}

Applying Bernoulli's theorem at points AA and BB:

PA+12ρvA2+ρgzA=PB+12ρvB2+ρgzBP_A + \dfrac{1}{2}\rho v_A^2 + \rho g z_A = P_B + \dfrac{1}{2}\rho v_B^2 + \rho g z_B

PA−PB=12ρ(vB2−vA2)+ρg(zB−zA)P_A - P_B = \dfrac{1}{2}\rho (v_B^2 - v_A^2) + \rho g (z_B - z_A)

Given zB−zA=h=1516 cm=1516×10−2 mz_B - z_A = h = \dfrac{15}{16} \text{ cm} = \dfrac{15}{16} \times 10^{-2} \text{ m}

12ρ(vB2−vA2)=12×1000×(0.52−0.252)=500×(0.25−0.0625)=500×0.1875=93.75 N/m2\dfrac{1}{2}\rho (v_B^2 - v_A^2) = \dfrac{1}{2} \times 1000 \times (0.5^2 - 0.25^2) = 500 \times (0.25 - 0.0625) = 500 \times 0.1875 = 93.75 \text{ N/m}^2

ρgh=1000×10×1516×10−2=100×1516=93.75 N/m2\rho g h = 1000 \times 10 \times \dfrac{15}{16} \times 10^{-2} = 100 \times \dfrac{15}{16} = 93.75 \text{ N/m}^2

PA−PB=93.75+93.75=187.5 N/m2≈188 N/m2P_A - P_B = 93.75 + 93.75 = 187.5 \text{ N/m}^2 \approx 188 \text{ N/m}^2
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