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MHT CET 2025 (25 Apr, Shift 2) · Bernoulli's Equation

Bernoulli's Equation question from MHT CET 2025 (25 Apr, Shift 2)

Q7MHT CET 2025 · 25 APR · SHIFT 2Medium

Bernoulli's Equation

A horizontal pipeline carries water in a streamline flow. At a point along the pipe, where the cross- sectional area is 10 cm210\text{ cm}^2, the velocity of water is 1 m/s1\text{ m/s} and pressure is 2000 Pa. The pressure of water at another point where the cross-sectional area 5 cm25\text{ cm}^2 is

[Given →\rightarrow density of water =1000 kg/m3= 1000\text{ kg/m}^3 ]

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Answer: (c)

  1. Continuity: v2=105×1=2 m/sv_2 = \dfrac{10}{5}\times 1 = 2\text{ m/s}
  2. Bernoulli: P2=P1−12ρ(v22−v12)=2000−12×1000×(4−1)P_2 = P_1 - \frac{1}{2}\rho (v_2^2 - v_1^2) = 2000 - \frac{1}{2}\times1000\times(4 - 1)
  3. P2=2000−1500=500 PaP_2 = 2000 - 1500 = 500\text{ Pa}
  4. So the answer is (c) 500 Pa.
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