Question:
Water flows in a horizontal pipe. At a point where speed is $2$ m/s, pressure is $2000$ Pa. At another point, speed becomes $4$ m/s. Find pressure at second point. (Density $= 1000$ kg/m$^3$)
Water flows in a horizontal pipe. At a point where speed is $2$ m/s, pressure is $2000$ Pa. At another point, speed becomes $4$ m/s. Find pressure at second point. (Density $= 1000$ kg/m$^3$)
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In a horizontal pipe, as the speed of the fluid increases, the pressure must decrease to satisfy Bernoulli's principle. Since the speed doubled, the kinetic energy term increased fourfold, leading to a significant drop in pressure.
Updated On: May 20, 2026
- $8000$ Pa
- $14000$ Pa
- $2000$ Pa
- $-4000$ Pa
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The Correct Option is D
Solution and Explanation
Concept:
For a fluid flowing through a horizontal pipe, Bernoulli's principle states that the sum of pressure energy and kinetic energy per unit volume remains constant.
\[ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \]
Where:
• $P$ is the pressure.
• $\rho$ is the density of the fluid.
• $v$ is the velocity of the fluid.
Step 1: Identify the given values.
From the question:
• Initial pressure ($P_1$) = $2000$ Pa
• Initial speed ($v_1$) = $2$ m/s
• Final speed ($v_2$) = $4$ m/s
• Density of water ($\rho$) = $1000$ kg/m$^3$
Step 2: Apply Bernoulli's equation.
Rearrange the formula to solve for $P_2$: \[ P_2 = P_1 + \frac{1}{2}\rho (v_1^2 - v_2^2) \] Substitute the values: \[ P_2 = 2000 + \frac{1}{2}(1000) (2^2 - 4^2) \] \[ P_2 = 2000 + 500 (4 - 16) \]
Step 3: Calculate the final pressure.
\[ P_2 = 2000 + 500 (-12) \] \[ P_2 = 2000 - 6000 \] \[ P_2 = -4000 \text{ Pa} \]
• $P$ is the pressure.
• $\rho$ is the density of the fluid.
• $v$ is the velocity of the fluid.
Step 1: Identify the given values.
From the question:
• Initial pressure ($P_1$) = $2000$ Pa
• Initial speed ($v_1$) = $2$ m/s
• Final speed ($v_2$) = $4$ m/s
• Density of water ($\rho$) = $1000$ kg/m$^3$
Step 2: Apply Bernoulli's equation.
Rearrange the formula to solve for $P_2$: \[ P_2 = P_1 + \frac{1}{2}\rho (v_1^2 - v_2^2) \] Substitute the values: \[ P_2 = 2000 + \frac{1}{2}(1000) (2^2 - 4^2) \] \[ P_2 = 2000 + 500 (4 - 16) \]
Step 3: Calculate the final pressure.
\[ P_2 = 2000 + 500 (-12) \] \[ P_2 = 2000 - 6000 \] \[ P_2 = -4000 \text{ Pa} \]
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