What is the pressure difference P_A−P_B in this fluid flow problem?

72 Pa. v_B=0.5m/s (from continuity). P_A−P_B=½×600×(0.25−0.01)=72 Pa.

0.5 m/s. From A_A×v_A=A_B×v_B: v_B=10⁻⁴×0.1/(20×10⁻⁶)=0.5 m/s.

Liquid density 600 kg/m³ A_A=1cm² A_B=20mm² v_A=10cm/s — pressure difference

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PhysicsFluid Mechanics

QMCQFluid Mechanics

A liquid of density 600 kg/m³ flows steadily in a tube of varying cross-section. $A_A=1.0\text{ cm}^2$, $A_B=20\text{ mm}^2$. Same horizontal level. Speed at $A=10$ cm/s. Pressure difference $P_A-P_B$ is ________ Pa.

A) 18 B) 144 C) 36 D) 72

✅ Correct Answer

D) 72 Pa

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Solution

Find v_B by continuity equation

$A_Av_A=A_Bv_B$
$v_B=\dfrac{1\times10^{-4}\times0.1}{20\times10^{-6}}=\dfrac{10^{-5}}{2\times10^{-5}}=0.5\text{ m/s}$

Apply Bernoulli's equation

$P_A-P_B=\tfrac{1}{2}\rho(v_B^2-v_A^2)=\tfrac{1}{2}\times600\times(0.25-0.01)=300\times0.24=\boxed{72\text{ Pa}}$

📘 Continuity + Bernoulli at Same Height

Continuity: $A_1v_1=A_2v_2$. Bernoulli at same height: $P_1-P_2=\frac{1}{2}\rho(v_2^2-v_1^2)$. Narrower section has higher speed and lower pressure.

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Important Concepts

Continuity EquationA_A×v_A=A_B×v_B. v_B=1e-4×0.1/(20e-6)=0.5 m/s. Narrower B means higher speed.

Bernoulli's PrincipleP+½ρv²=const at same height. Higher speed→lower pressure. P_A−P_B=½ρ(v_B²−v_A²).

Unit Conversion1cm²=10⁻⁴m², 20mm²=20×10⁻⁶m², 10cm/s=0.1m/s. Always work in SI.

Numerical ResultP_A−P_B=300×(0.5²−0.1²)=300×(0.25−0.01)=300×0.24=72 Pa.

FAQs

Why is pressure lower at B?

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v_B>v_A so by Bernoulli: P_B

What is v_B?

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v_B=A_A×v_A/A_B=10⁻⁴×0.1/(20×10⁻⁶)=0.5 m/s.

Why do heights cancel?

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A and B are at the same horizontal level, so ρgh_A=ρgh_B in Bernoulli's equation.

What application uses this?

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Venturimeters measure flow rate using the pressure difference between wide and narrow sections.

Is this from JEE Main 2026?

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Yes, this appeared in JEE Main 2026.

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