Question:
A liquid of density 600 kg/m³ flowing steadily in a tube of varying cross-section. The cross-section at a point A is 1.0 cm² and that at B is 20 mm². Both the points A and B are in same horizontal plane, the speed of the liquid at A is 10 cm/s. The difference in pressures at A and B points is ________ Pa.
A liquid of density 600 kg/m³ flowing steadily in a tube of varying cross-section. The cross-section at a point A is 1.0 cm² and that at B is 20 mm². Both the points A and B are in same horizontal plane, the speed of the liquid at A is 10 cm/s. The difference in pressures at A and B points is ________ Pa.
Updated On: Apr 13, 2026
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
We apply the Principle of Continuity to find the velocity at point B and then use Bernoulli's Equation for a horizontal flow to find the pressure difference.
Step 2: Key Formula or Approach:
1. Continuity: \(A_1 v_1 = A_2 v_2\).
2. Bernoulli: \(P_A + \frac{1}{2}\rho v_A^2 = P_B + \frac{1}{2}\rho v_B^2\).
Step 3: Detailed Explanation:
Convert all units to SI: \(A_A = 1.0 \text{ cm}^2 = 10^{-4} \text{ m}^2\) \(A_B = 20 \text{ mm}^2 = 20 \times 10^{-6} \text{ m}^2 = 2 \times 10^{-5} \text{ m}^2\) \(v_A = 10 \text{ cm/s} = 0.1 \text{ m/s}\), \(\rho = 600 \text{ kg/m}^3\) Find \(v_B\): \[ (10^{-4})(0.1) = (2 \times 10^{-5}) v_B \implies 10^{-5} = 2 \times 10^{-5} v_B \implies v_B = 0.5 \text{ m/s} \] Pressure difference \(\Delta P = P_A - P_B = \frac{1}{2}\rho (v_B^2 - v_A^2)\): \[ \Delta P = \frac{1}{2}(600) (0.5^2 - 0.1^2) = 300 (0.25 - 0.01) \] \[ \Delta P = 300 (0.24) = 72 \text{ Pa} \]
Step 4: Final Answer:
The difference in pressures is 72 Pa.
We apply the Principle of Continuity to find the velocity at point B and then use Bernoulli's Equation for a horizontal flow to find the pressure difference.
Step 2: Key Formula or Approach:
1. Continuity: \(A_1 v_1 = A_2 v_2\).
2. Bernoulli: \(P_A + \frac{1}{2}\rho v_A^2 = P_B + \frac{1}{2}\rho v_B^2\).
Step 3: Detailed Explanation:
Convert all units to SI: \(A_A = 1.0 \text{ cm}^2 = 10^{-4} \text{ m}^2\) \(A_B = 20 \text{ mm}^2 = 20 \times 10^{-6} \text{ m}^2 = 2 \times 10^{-5} \text{ m}^2\) \(v_A = 10 \text{ cm/s} = 0.1 \text{ m/s}\), \(\rho = 600 \text{ kg/m}^3\) Find \(v_B\): \[ (10^{-4})(0.1) = (2 \times 10^{-5}) v_B \implies 10^{-5} = 2 \times 10^{-5} v_B \implies v_B = 0.5 \text{ m/s} \] Pressure difference \(\Delta P = P_A - P_B = \frac{1}{2}\rho (v_B^2 - v_A^2)\): \[ \Delta P = \frac{1}{2}(600) (0.5^2 - 0.1^2) = 300 (0.25 - 0.01) \] \[ \Delta P = 300 (0.24) = 72 \text{ Pa} \]
Step 4: Final Answer:
The difference in pressures is 72 Pa.
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