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

To solve for the cross-sectional area at point B, we can use the principle of conservation of mass for an incompressible fluid, which is expressed by the equation of continuity:

\( A_1 V_1 = A_2 V_2 \)

where:

• \( A_1 \) is the cross-sectional area at point A.
• \( V_1 \) is the velocity of the fluid at point A.
• \( A_2 \) is the cross-sectional area at point B.
• \( V_2 \) is the velocity of the fluid at point B.

Given:

• \( A_1 = 1 \, \text{m}^2 \)
• \( V_1 = 2 \, \text{m/s} \)
• \( V_2 = 4 \, \text{m/s} \)

We need to find \( A_2 \).

Substituting the given values into the continuity equation, we have:

\( 1 \cdot 2 = A_2 \cdot 4 \)

\( 2 = 4A_2 \)

Solving for \( A_2 \):

\( A_2 = \frac{2}{4} \)

\( A_2 = 0.5 \, \text{m}^2 \)

The cross-sectional area at point B is therefore \( 0.5 \, \text{m}^2 \).