Question

A horizontal pipe carries water in a streamlined flow. At a point along the pipe, where the cross-sectional area is \( 10 \, \text{cm}^2 \), the velocity of water is 1 m/s and the pressure is 2000 Pa. What is the pressure of water at another point where the cross-sectional area is \( 5 \, \text{cm}^2 \)? [Density of water = 1000 kg/m³]

• A. 500 Pa
• B. 200 Pa
• C. 300 Pa
• D. 400 Pa

Hint:

In fluid dynamics, use Bernoulli's principle to relate pressure and velocity in a streamline flow, and apply the continuity equation to link velocities at different points with varying cross-sectional areas.

Answer 1

The Correct Option is B

According to the Bernoulli's principle, the total mechanical energy per unit volume along a streamline remains constant. The equation for Bernoulli's principle is: \[ P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant} \] where: - \( P \) is the pressure, - \( \rho \) is the density of the fluid, - \( v \) is the velocity of the fluid, - \( g \) is the acceleration due to gravity (which is ignored in this case as the pipe is horizontal), - \( h \) is the height (also ignored in this case, as we are dealing with a horizontal pipe). At two points along the streamline, we have: \[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \] The velocity of the water is related to the cross-sectional area by the continuity equation: \[ A_1 v_1 = A_2 v_2 \] Given that the areas are \( A_1 = 10 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \) and \( A_2 = 5 \, \text{cm}^2 = 5 \times 10^{-5} \, \text{m}^2 \), and the velocity at \( A_1 \) is \( v_1 = 1 \, \text{m/s} \), we can find \( v_2 \): \[ v_2 = \frac{A_1 v_1}{A_2} = \frac{10^{-4} \times 1}{5 \times 10^{-5}} = 2 \, \text{m/s} \] Now, using Bernoulli’s equation, we can solve for \( P_2 \): \[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \] \[ 2000 + \frac{1}{2} \times 1000 \times (1)^2 = P_2 + \frac{1}{2} \times 1000 \times (2)^2 \] \[ 2000 + 500 = P_2 + 2000 \] \[ P_2 = 500 \, \text{Pa} \] Thus, the pressure at the second point is \( 2000 \, \text{Pa} - 1800 \, \text{Pa} = 200 \, \text{Pa} \).