Question

A fluid of density \( 800 \, \text{kg/m}^3 \) is flowing through a pipe of varying cross-sectional area. The velocity of the fluid at point A is \( 2 \, \text{m/s} \), and the velocity at point B is \( 4 \, \text{m/s} \). If the cross-sectional area at point A is \( 1 \, \text{m}^2 \), find the cross-sectional area at point B.

• A. \( 0.5 \, \text{m}^2 \)
• B. \( 1.5 \, \text{m}^2 \)
• C. \( 2.0 \, \text{m}^2 \)
• D. \( 4.0 \, \text{m}^2 \)

Hint:

In fluid dynamics, the principle of continuity ensures that the mass flow rate remains constant in an incompressible fluid. This means that if the velocity of the fluid increases, the cross-sectional area must decrease.

Answer 1

The Correct Option is A

Using the principle of continuity for incompressible fluids, the mass flow rate must remain constant. This is given by the equation: \[ A_1 v_1 = A_2 v_2 \] Where: - \( A_1 = 1 \, \text{m}^2 \) is the cross-sectional area at point A, - \( v_1 = 2 \, \text{m/s} \) is the velocity at point A, - \( A_2 \) is the cross-sectional area at point B, - \( v_2 = 4 \, \text{m/s} \) is the velocity at point B. Substitute the known values: \[ 1 \times 2 = A_2 \times 4 \] \[ A_2 = \frac{2}{4} = 0.5 \, \text{m}^2 \] Thus, the cross-sectional area at point B is \( 0.5 \, \text{m}^2 \).