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NEET
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Physics
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Mechanical Properties of Fluids

NEET · Physics · Mechanical Properties of Fluids

Oil of density 800 kg m^ -3 flows steadily through a horizontal...

Oil of density 800 kg m−3800 \text{ kg m}^{-3} flows steadily through a horizontal pipe. The cross-sectional areas of the pipe at two sections A and B are 40 cm240 \text{ cm}^2 and 10 cm210 \text{ cm}^2 respectively. If the volume flow rate of the oil is 2000 cm3 s−12000 \text{ cm}^3 \text{ s}^{-1}, what is the pressure difference between the two sections?

Options

  1. A900 Pa900 \text{ Pa}
  2. B1500 Pa1500 \text{ Pa}
  3. C1700 Pa1700 \text{ Pa}
  4. D3000 Pa3000 \text{ Pa}

Correct answer

B. 1500 Pa1500 \text{ Pa}

Step-by-step solution

From the given volume flow rate QQ, the velocity of oil at section A is:

vA=QAA=2000 cm3 s−140 cm2=50 cm s−1=0.5 m s−1v_A = \dfrac{Q}{A_A} = \dfrac{2000 \text{ cm}^3 \text{ s}^{-1}}{40 \text{ cm}^2} = 50 \text{ cm s}^{-1} = 0.5 \text{ m s}^{-1}

The velocity at section B is:

vB=QAB=2000 cm3 s−110 cm2=200 cm s−1=2 m s−1v_B = \dfrac{Q}{A_B} = \dfrac{2000 \text{ cm}^3 \text{ s}^{-1}}{10 \text{ cm}^2} = 200 \text{ cm s}^{-1} = 2 \text{ m s}^{-1}

According to Bernoulli's principle for a horizontal pipe:

PA+12ρvA2=PB+12ρvB2P_A + \dfrac{1}{2}\rho v_A^2 = P_B + \dfrac{1}{2}\rho v_B^2

The pressure difference is:

ΔP=PA−PB=12ρ(vB2−vA2)\Delta P = P_A - P_B = \dfrac{1}{2}\rho (v_B^2 - v_A^2)

Substitute the values:

ΔP=12×800×(22−0.52)\Delta P = \dfrac{1}{2} \times 800 \times (2^2 - 0.5^2)

ΔP=400×(4−0.25)=400×3.75=1500 Pa\Delta P = 400 \times (4 - 0.25) = 400 \times 3.75 = 1500 \text{ Pa}

Answer: 1500 Pa1500 \text{ Pa}
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