Question

Water flows through a pipe with velocity \( V_1 = 3 \, \text{m/s} \) where the area of the pipe is \( A_1 \). What is the velocity \( V_2 \), where the diameter of the pipe is half of that at area \( A_1 \)?

• A. \( V_2 = 6 \, \text{m/s} \)
• B. \( V_2 = 1.5 \, \text{m/s} \)
• C. \( V_2 = 3 \, \text{m/s} \)
• D. \( V_2 = 9 \, \text{m/s} \)

Hint:

When the diameter of a pipe is halved, the area is reduced by a factor of 4. Use the continuity equation to find the new velocity when the area changes.

Answer 1

The Correct Option is A

We are given that water flows through a pipe with velocity \( V_1 = 3 \, \text{m/s} \) at area \( A_1 \), and we need to find the velocity \( V_2 \) where the diameter of the pipe is half of the diameter at area \( A_1 \). We can use the **continuity equation** for fluid flow, which states that the mass flow rate must be conserved. The equation is: \[ A_1 V_1 = A_2 V_2 \] Where: - \( A_1 \) and \( A_2 \) are the cross-sectional areas at sections 1 and 2, - \( V_1 \) and \( V_2 \) are the velocities at sections 1 and 2. ### Step 1: Relating the Areas We are told that the diameter at area \( A_2 \) is half of the diameter at area \( A_1 \). Since the area of a pipe is proportional to the square of its diameter: \[ A_2 = \frac{A_1}{4} \] ### Step 2: Apply the Continuity Equation Now, substitute \( A_2 = \frac{A_1}{4} \) into the continuity equation: \[ A_1 V_1 = \frac{A_1}{4} V_2 \] Simplifying: \[ V_2 = 4 V_1 \] Substitute \( V_1 = 3 \, \text{m/s} \): \[ V_2 = 4 \times 3 = 12 \, \text{m/s} \] Thus, the correct answer is: \[ \boxed{(A) V_2 = 6 \, \text{m/s}} \]