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. 18
• B. 144
• C. 36
• D. 72

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The problem requires finding the pressure difference between two points in a steadily flowing liquid. Because the tube is horizontal, there is no change in potential energy. We apply the Principle of Continuity to find the velocity at point B, and then use Bernoulli's equation to find the pressure difference.
Step 2: Key Formula or Approach:
1. Equation of Continuity: $A_A v_A = A_B v_B$.
2. Bernoulli's Equation for horizontal flow: $P_A + \frac{1}{2}\rho v_A^2 = P_B + \frac{1}{2}\rho v_B^2$.
Step 3: Detailed Explanation:
Given values:
Density $\rho = 600\text{ kg/m}^3$.
Area at A, $A_A = 1.0\text{ cm}^2 = 1.0 \times 10^{-4}\text{ m}^2$.
Area at B, $A_B = 20\text{ mm}^2 = 20 \times 10^{-6}\text{ m}^2 = 0.2 \times 10^{-4}\text{ m}^2$.
Velocity at A, $v_A = 10\text{ cm/s} = 0.1\text{ m/s}$.
Using the equation of continuity to find velocity at B ($v_B$):
$A_A v_A = A_B v_B \implies (1.0 \times 10^{-4})(0.1) = (0.2 \times 10^{-4})v_B$.
$0.1 = 0.2 v_B \implies v_B = \frac{0.1}{0.2} = 0.5\text{ m/s}$.
Now apply Bernoulli's equation. Since both points are on the same horizontal plane, $h_A = h_B$, so the $\rho g h$ terms cancel out.
$P_A + \frac{1}{2}\rho v_A^2 = P_B + \frac{1}{2}\rho v_B^2$.
The pressure difference $\Delta P$ is:
$P_A - P_B = \frac{1}{2}\rho v_B^2 - \frac{1}{2}\rho v_A^2 = \frac{1}{2}\rho (v_B^2 - v_A^2)$.
Substitute the known values:
$P_A - P_B = \frac{1}{2} (600) \left[ (0.5)^2 - (0.1)^2 \right]$.
$= 300 \left[ 0.25 - 0.01 \right]$.
$= 300 \times 0.24 = 72\text{ Pa}$.
Step 4: Final Answer:
The difference in pressures is 72 Pa.