The Question is from neet-physics mechanical-properties-of-fluids , and it was created by the author VENKATESWARARAO

Question

A non-viscous, incompressible fluid flows steadily through a horizontal pipe with a varying cross-sectional area. At a point where the cross-sectional area is $A_1$, the fluid velocity is $v_1$ and the pressure is $P_1$. At another point where the area is $A_2 = A_1/2$, the velocity is $v_2$ and the pressure is $P_2$. If $\rho$ is the fluid density, which of the following expressions correctly relates $P_1$ and $P_2$?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
 **Primary Concept**: Bernoulli's equation for fluid flow
 **Related Topics**: Fluid dynamics, conservation of mass, continuity equation
 **Prerequisites**: Understanding of pressure, velocity, and density in fluid flow; knowledge of Bernoulli's equation.
## SOLUTION PATHWAY
**Given Information**:
  $A_2 = A_1/2$
  Fluid is non-viscous and incompressible
  Flow is steady
  Pipe is horizontal
  $A_1$, $v_1$, $P_1$ at point 1
  $A_2$, $v_2$, $P_2$ at point 2
**Method & Steps**:
 1. Apply the continuity equation (conservation of mass): $A_1v_1 = A_2v_2$. Since $A_2 = A_1/2$, we have $v_2 = 2v_1$.
  Reasoning: The mass flow rate must remain constant in a steady, incompressible flow.
  Formula/Rule: $A_1v_1 = A_2v_2$
 2. Apply Bernoulli's equation for a horizontal pipe (no change in height): $P_1 + \frac{1}{2} ho v_1^2 = P_2 + \frac{1}{2} ho v_2^2$.
  Working: Substitute $v_2 = 2v_1$ into Bernoulli's equation: $P_1 + \frac{1}{2} ho v_1^2 = P_2 + \frac{1}{2} ho (2v_1)^2 = P_2 + 2 ho v_1^2$
  Key Point: Bernoulli's equation states that the sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline.
 3. Rearrange the equation to solve for $P_1$ in terms of $P_2$ and $v_1$: $P_1 = P_2 + \frac{3}{2} ho v_1^2$
  Result: $P_1 = P_2 + \frac{3}{2} ho v_1^2$
  Verification: The equation shows that the pressure $P_1$ is higher than $P_2$ because the velocity is lower at point 1, consistent with Bernoulli's principle.
 ## OPTION ANALYSIS
A :x: Incorrect. This option only accounts for the kinetic energy difference, neglecting the factor of 4 from the velocity change.
B ✓: Correct. This option correctly reflects the pressure difference based on Bernoulli's equation and the continuity equation.
C :x: Incorrect. This option overestimates the pressure difference by a factor of 2.
D :x: Incorrect. This option uses $v_2$ instead of $v_1$ which is not consistent with the question.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Forgetting to apply the continuity equation to find the relationship between $v_1$ and $v_2$: Remember that for incompressible fluids, the mass flow rate is constant.
  2.Incorrectly applying Bernoulli's equation: Ensure you use the correct form for horizontal flow, and account for the velocity difference.
 **Quick Tips**:
  1.Remember the continuity equation ($A_1v_1 = A_2v_2$) and Bernoulli's equation ($P_1 + \frac{1}{2}ho v_1^2 + ho gh_1 = P_2 + \frac{1}{2}ho v_2^2 + ho gh_2$) as fundamental tools for solving fluid dynamics problems.
  2.Visualize the flow:  A smaller cross-sectional area leads to a higher velocity and lower pressure, and vice versa.

VENKATESWARARAO · Answer

$P_1 = P_2 + \frac{1}{2} \rho v_1^2$

VENKATESWARARAO · Answer

The correct Option is "$P_1 = P_2 + \frac{3}{2} \rho v_1^2$"

VENKATESWARARAO · Answer

$P_1 = P_2 + 2 \rho v_1^2$

VENKATESWARARAO · Answer

$P_1 = P_2 + \frac{1}{2} \rho v_2^2$

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