Question

Water is flowing in a streamline manner in a horizontal pipe. If the pressure at a point where cross-sectional area is \( 10 \) cm\(^2\) and velocity \( 1 \) m/s is \( 2000 \) Pa, then the pressure of water at another point where the cross-sectional area is \( 5 \) cm\(^2\) is:

• A. \( 2500 \) Pa
• B. \( 2000 \) Pa
• C. \( 1000 \) Pa
• D. \( 500 \) Pa

Hint:

For fluid flow in a horizontal pipe, use Bernoulli’s equation: \[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2. \] Also, apply the continuity equation \( A_1 v_1 = A_2 v_2 \) to find velocity at different cross-sections.

Answer 1

The Correct Option is D

Step 1: Applying the Continuity Equation The continuity equation states that for an incompressible fluid: \[ A_1 v_1 = A_2 v_2. \] Given: \[ A_1 = 10 \text{ cm}^2 = 10 \times 10^{-4} \text{ m}^2, \quad A_2 = 5 \text{ cm}^2 = 5 \times 10^{-4} \text{ m}^2. \] \[ v_1 = 1 \text{ m/s}. \] Using the continuity equation: \[ 10 \times 10^{-4} \times 1 = 5 \times 10^{-4} \times v_2. \] Solving for \( v_2 \): \[ v_2 = \frac{10 \times 10^{-4} \times 1}{5 \times 10^{-4}} = 2 \text{ m/s}. \]
Step 2: Applying Bernoulli’s Equation Bernoulli’s equation states: \[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2. \] Given: \[ P_1 = 2000 \text{ Pa}, \quad v_1 = 1 \text{ m/s}, \quad v_2 = 2 \text{ m/s}. \] Assuming water density \( \rho = 1000 \) kg/m\(^3\), we substitute values: \[ 2000 + \frac{1}{2} (1000) (1)^2 = P_2 + \frac{1}{2} (1000) (2)^2. \] \[ 2000 + 500 = P_2 + 2000. \] \[ P_2 = 500 \text{ Pa}. \] Thus, the correct answer is: \[ \boxed{500 \text{ Pa}}. \]

Answer 2

Step 1: Applying the Continuity Equation

The continuity equation for an incompressible fluid (such as water) is expressed as:
\[ A_1 v_1 = A_2 v_2 \]
where:
- \( A_1 \) and \( A_2 \) are the cross-sectional areas of the pipe at sections 1 and 2 respectively
- \( v_1 \) and \( v_2 \) are the fluid velocities at those sections

Given values:
\[ A_1 = 10 \text{ cm}^2 = 10 \times 10^{-4} \text{ m}^2 \]
\[ A_2 = 5 \text{ cm}^2 = 5 \times 10^{-4} \text{ m}^2 \]
\[ v_1 = 1 \text{ m/s} \]

Plugging into the continuity equation:
\[ 10 \times 10^{-4} \times 1 = 5 \times 10^{-4} \times v_2 \]
Solving for \( v_2 \):
\[ v_2 = \frac{10 \times 10^{-4}}{5 \times 10^{-4}} = 2 \text{ m/s} \]

Step 2: Applying Bernoulli’s Equation

Bernoulli’s equation relates the pressure and velocity at two points along a streamline in a steady, incompressible, non-viscous fluid:
\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \]
Given:
\[ P_1 = 2000 \text{ Pa}, \quad v_1 = 1 \text{ m/s}, \quad v_2 = 2 \text{ m/s}, \quad \rho = 1000 \text{ kg/m}^3 \]
Substituting into Bernoulli's equation:
\[ 2000 + \frac{1}{2}(1000)(1)^2 = P_2 + \frac{1}{2}(1000)(2)^2 \]
\[ 2000 + 500 = P_2 + 2000 \]
Solving for \( P_2 \):
\[ P_2 = 2500 - 2000 = 500 \text{ Pa} \]

Final Answer:
\[ \boxed{500 \text{ Pa}} \]
This is the pressure at the narrower section of the pipe where the velocity increases due to the decrease in area.