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

Question

A viscous liquid flows through a horizontal pipe of varying cross-section.  At point A, the radius is 'r' and the velocity is 'v'. At point B, the radius is 'r/2'. Ignoring any energy losses due to viscosity, what is the pressure difference between points A and B?

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
 **Primary Concept**: Bernoulli's principle for fluid flow in a horizontal pipe.
 **Related Topics**: Fluid dynamics, conservation of energy, continuity equation.
 **Prerequisites**: Understanding of pressure, velocity, and area relationships in fluid flow; knowledge of Bernoulli's equation.
## SOLUTION PATHWAY
**Given Information**:
  $r_A = r$, $v_A = v$
  $r_B = r/2$
  Energy losses due to viscosity are negligible.
  Horizontal pipe (no change in height).
**Method & Steps**:
 1. Apply the continuity equation to find the velocity at point B.
  Reasoning: The mass flow rate must be constant throughout the pipe.
  Formula/Rule: $A_A v_A = A_B v_B$, where A is the cross-sectional area ($A = \pi r^2$).
 2. Apply Bernoulli's equation to relate pressure and velocity at points A and B.
  Working: Bernoulli's equation states that $P_A + \frac{1}{2}ho v_A^2 = P_B + \frac{1}{2}ho v_B^2$  for a horizontal pipe where $ho$ is the density of the liquid.  Substitute the velocity found in step 1.
  Key Point: Since the pipe is horizontal, the potential energy terms cancel out.
 3. Solve for the pressure difference ($P_A - P_B$).
  Result:  From the continuity equation, $\pi r^2 v = \pi (r/2)^2 v_B$, which gives $v_B = 4v$.  Substituting into Bernoulli's equation:
       $P_A - P_B = \frac{1}{2}ho (v_B^2 - v_A^2) = \frac{1}{2}ho ((4v)^2 - v^2) = \frac{1}{2}ho (16v^2 - v^2) = \frac{15}{2}ho v^2$
  Verification: The units are consistent ($\frac{kg}{m^3} \cdot \frac{m^2}{s^2} = \frac{kg}{m \cdot s^2} = Pa$, which are the units of pressure).
 ## OPTION ANALYSIS
A :x: Incorrect calculation of velocity at point B or incorrect application of Bernoulli's equation.
B :x: Incorrect calculation of velocity at point B or incorrect application of Bernoulli's equation.
C ✓: Correct application of continuity equation and Bernoulli's equation.
D :x: This would only be true if the velocity at points A and B were equal, which is not the case due to the change in cross-sectional area.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Forgetting to apply the continuity equation to relate velocities at different points in the pipe.
  2.Incorrectly applying Bernoulli's equation, neglecting terms or making algebraic errors.
 **Quick Tips**:
  1.Always start with the continuity equation in problems involving fluid flow in pipes with varying cross-sections.
  2.Remember that Bernoulli's equation applies only when viscous effects are negligible.

SUBHANI · Answer

$\frac{3 \rho v^2}{2}$

SUBHANI · Answer

$\frac{7 \rho v^2}{2}$

SUBHANI · Answer

The correct Option is "$\frac{15 \rho v^2}{2}$"

SUBHANI · Answer

$0$

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