Question

A fluid flows through a pipe with a varying cross-sectional area. If the velocity of the fluid is \( v_1 = 4 \, \text{m/s} \) at a point where the cross-sectional area is \( A_1 = 2 \, \text{m}^2 \), and the velocity at another point where the cross-sectional area is \( A_2 = 1 \, \text{m}^2 \) is \( v_2 \), what is the velocity \( v_2 \)?

• A. \( 8 \, \text{m/s} \)
• B. \( 4 \, \text{m/s} \)
• C. \( 2 \, \text{m/s} \)
• D. \( 1 \, \text{m/s} \)

Hint:

For fluids flowing through a pipe with varying cross-section, use the principle of continuity \( A_1 v_1 = A_2 v_2 \) to find the velocity at different points in the pipe.

Answer 1

The Correct Option is A

We are given a fluid flowing through a pipe with varying cross-sectional areas. The velocity of the fluid is \( v_1 = 4 \, \text{m/s} \) at a point where the cross-sectional area is \( A_1 = 2 \, \text{m}^2 \), and the velocity at another point is \( v_2 \), where the cross-sectional area is \( A_2 = 1 \, \text{m}^2 \). We need to find \( v_2 \). Step 1: Use the principle of continuity The principle of continuity for fluid flow states that the mass flow rate must be constant throughout the pipe. For an incompressible fluid, this means that the product of the cross-sectional area and the velocity at any point in the pipe is constant: \[ A_1 v_1 = A_2 v_2 \] Step 2: Substitute the known values Substitute \( A_1 = 2 \, \text{m}^2 \), \( v_1 = 4 \, \text{m/s} \), and \( A_2 = 1 \, \text{m}^2 \) into the equation: \[ 2 \times 4 = 1 \times v_2 \] \[ 8 = v_2 \] Answer: The velocity \( v_2 \) is \( 8 \, \text{m/s} \), so the correct answer is option (1).