Question

A horizontal pipeline carries water in streamline flow. At $A_1 = 10$ cm$^2$, $v_1 = 1$ m/s and $P_1 = 2000$ Pa. The pressure at $A_2 = 5$ cm$^2$ is \dots [$\rho = 1000$ kg/m$^3$]

• A. 1000 Pa
• B. 750 Pa
• C. 500 Pa
• D. 250 Pa

Hint:

This is the Venturi effect in action! Wherever the pipe gets narrower, the fluid is forced to speed up (Continuity). As its kinetic energy increases, its static pressure must simultaneously drop to conserve total energy (Bernoulli).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Water flows through a horizontal pipe that undergoes a constriction (the cross-sectional area halves). We must use fluid dynamics principles to find the water pressure at this narrower section.

Step 2: Key Formula or Approach:
1. Equation of Continuity: To find the velocity at the second point.
$$A_1 v_1 = A_2 v_2$$
2. Bernoulli's Equation: For a strictly horizontal pipe ($h_1 = h_2$), the potential energy terms cancel out.
$$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$

Step 3: Detailed Explanation:
First, find the velocity $v_2$ at the constriction:
$$10 \text{ cm}^2 \times 1 \text{ m/s} = 5 \text{ cm}^2 \times v_2$$
(We don't need to convert cm$^2$ to m$^2$ because the unit ratio perfectly cancels out).
$$10 = 5 v_2 \implies v_2 = 2 \text{ m/s}$$
Next, apply Bernoulli's equation to find $P_2$:
$$P_2 = P_1 + \frac{1}{2}\rho v_1^2 - \frac{1}{2}\rho v_2^2$$
$$P_2 = P_1 + \frac{1}{2}\rho (v_1^2 - v_2^2)$$
Substitute the given values ($P_1 = 2000 \text{ Pa}$, $\rho = 1000 \text{ kg/m}^3$, $v_1 = 1 \text{ m/s}$, $v_2 = 2 \text{ m/s}$):
$$P_2 = 2000 + \frac{1}{2}(1000) (1^2 - 2^2)$$
$$P_2 = 2000 + 500 (1 - 4)$$
$$P_2 = 2000 + 500 (-3)$$
$$P_2 = 2000 - 1500$$
$$P_2 = 500 \text{ Pa}$$

Step 4: Final Answer:
The pressure at the narrower section drops to 500 Pa, matching option (c).