Question:
Water flows in a horizontal pipe. At a point where speed is \(2\,\text{m/s}\), pressure is \(2000\,\text{Pa}\). At another point, speed becomes \(4\,\text{m/s}\). Find pressure at second point. (Density \(= 1000\,\text{kg/m}^3\))
Water flows in a horizontal pipe. At a point where speed is \(2\,\text{m/s}\), pressure is \(2000\,\text{Pa}\). At another point, speed becomes \(4\,\text{m/s}\). Find pressure at second point. (Density \(= 1000\,\text{kg/m}^3\))
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According to the Venturi effect (derived from Bernoulli's principle), as the speed of a flowing fluid increases, its static pressure must decrease. Since the speed doubled, we should expect a substantial decrease in pressure.
Updated On: May 24, 2026
- $8000\text{ Pa}$
- $14000\text{ Pa}$
- $2000\text{ Pa}$
- $-4000\text{ Pa}$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
The question asks to find the fluid pressure at a second point in a horizontal pipe, given the fluid velocity and pressure at the first point, and the velocity at the second point.
Step 2: Key Formula or Approach:
For steady, incompressible, and non-viscous fluid flow along a horizontal pipe, the height component remains constant ($h_1 = h_2$). We apply Bernoulli's Theorem:
\[ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \] where $P$ represents static pressure, $\rho$ is the fluid density, and $v$ is the flow velocity at the respective points.
Step 3: Detailed Explanation:
• Let us list the given parameters:
Initial pressure, $P_1 = 2000\text{ Pa}$
Initial speed, $v_1 = 2\text{ m/s}$
Final speed, $v_2 = 4\text{ m/s}$
Density of water, $\rho = 1000\text{ kg/m}^3$
• We rearrange Bernoulli's equation to solve for the final pressure, $P_2$:
\[ P_2 = P_1 + \frac{1}{2}\rho v_1^2 - \frac{1}{2}\rho v_2^2 \] \[ P_2 = P_1 + \frac{1}{2}\rho (v_1^2 - v_2^2) \]
• Substituting the values into this equation:
\[ P_2 = 2000 + \frac{1}{2}(1000)(2^2 - 4^2) \] \[ P_2 = 2000 + 500(4 - 16) \] \[ P_2 = 2000 + 500(-12) \] \[ P_2 = 2000 - 6000 \] \[ P_2 = -4000\text{ Pa} \]
• The calculated pressure is $-4000\text{ Pa}$. In fluid mechanics, a negative gauge pressure represents a partial vacuum, meaning the local pressure is lower than the reference atmospheric pressure.
Step 4: Final Answer:
The pressure at the second point in the pipe is $-4000\text{ Pa}$.
The question asks to find the fluid pressure at a second point in a horizontal pipe, given the fluid velocity and pressure at the first point, and the velocity at the second point.
Step 2: Key Formula or Approach:
For steady, incompressible, and non-viscous fluid flow along a horizontal pipe, the height component remains constant ($h_1 = h_2$). We apply Bernoulli's Theorem:
\[ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \] where $P$ represents static pressure, $\rho$ is the fluid density, and $v$ is the flow velocity at the respective points.
Step 3: Detailed Explanation:
• Let us list the given parameters:
Initial pressure, $P_1 = 2000\text{ Pa}$
Initial speed, $v_1 = 2\text{ m/s}$
Final speed, $v_2 = 4\text{ m/s}$
Density of water, $\rho = 1000\text{ kg/m}^3$
• We rearrange Bernoulli's equation to solve for the final pressure, $P_2$:
\[ P_2 = P_1 + \frac{1}{2}\rho v_1^2 - \frac{1}{2}\rho v_2^2 \] \[ P_2 = P_1 + \frac{1}{2}\rho (v_1^2 - v_2^2) \]
• Substituting the values into this equation:
\[ P_2 = 2000 + \frac{1}{2}(1000)(2^2 - 4^2) \] \[ P_2 = 2000 + 500(4 - 16) \] \[ P_2 = 2000 + 500(-12) \] \[ P_2 = 2000 - 6000 \] \[ P_2 = -4000\text{ Pa} \]
• The calculated pressure is $-4000\text{ Pa}$. In fluid mechanics, a negative gauge pressure represents a partial vacuum, meaning the local pressure is lower than the reference atmospheric pressure.
Step 4: Final Answer:
The pressure at the second point in the pipe is $-4000\text{ Pa}$.
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