The Question is from neet-physics mechanical-properties-of-fluids , and it was created by the author Mr. ANIL KUMAR.G

Question

A capillary tube of radius 'r' is dipped vertically in a liquid of density '$ρ$' and surface tension 'T'. If the angle of contact is $	heta$, the pressure difference between the two sides of the curved surface at the top of the meniscus is:

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
 **Primary Concept**: Capillary action and pressure difference across a curved meniscus.
 **Related Topics**: Surface tension, contact angle, Young-Laplace equation.
 **Prerequisites**: Understanding of surface tension, pressure, and contact angle.
## SOLUTION PATHWAY
**Given Information**:
  Radius of capillary tube = r
  Density of liquid = $ho$
  Surface tension of liquid = T
  Contact angle = $	heta$
**Method & Steps**:
 1. The pressure difference across a curved surface (meniscus) in a liquid is given by the Young-Laplace equation.  For a spherical meniscus (a good approximation for a small radius capillary), the equation is:
  Reasoning: The pressure inside the meniscus is less than the pressure outside.  The difference is proportional to the surface tension and inversely proportional to the radius of curvature.
  Formula/Rule: $\Delta P = \frac{2T}{R}$, where $\Delta P$ is the pressure difference, T is the surface tension, and R is the radius of curvature.
 2. In a capillary tube, the radius of curvature of the meniscus (R) is related to the radius of the tube (r) and the contact angle ($	heta$) by the equation:
  Working: $R = \frac{r}{\cos 	heta}$
  Key Point: The radius of curvature is always positive, and the cosine of the contact angle accounts for the shape of the meniscus (concave or convex).
 3. Substitute the expression for R into the Young-Laplace equation:
  Result: $\Delta P = \frac{2T}{r/\cos	heta} = \frac{2T \cos 	heta}{r}$
  Verification: The units are consistent (N/m²/m = N/m² = Pa, which is a unit of pressure).
 ## OPTION ANALYSIS
[A] :x: Incorrect. It misses the factor of 2 from the Young-Laplace equation.
[B] ✓: Correct. This is the correct application of the Young-Laplace equation considering the relationship between the radius of curvature and the contact angle.
[C] :x: Incorrect. It uses sine instead of cosine, and misses the factor of 2.
[D] :x: Incorrect. It uses sine instead of cosine.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Using the wrong trigonometric function (sine instead of cosine): Pay close attention to the geometry of the meniscus and the definition of the contact angle.
  2.Forgetting the factor of 2 in the Young-Laplace equation: This factor arises from the curvature in two dimensions.
 **Quick Tips**:
  1.Remember the Young-Laplace equation is a fundamental relationship in fluid mechanics.
  2.Draw a diagram to understand the geometry and the relationship between the radius of curvature, the tube radius, and the contact angle.

Mr. ANIL KUMAR.G · Answer

$ \frac{T \cos 	heta}{r}$

Mr. ANIL KUMAR.G · Answer

The correct Option is "$ \frac{2T \cos 	heta}{r}$"

Mr. ANIL KUMAR.G · Answer

$ \frac{T \sin 	heta}{r}$

Mr. ANIL KUMAR.G · Answer

$ \frac{2T \sin 	heta}{r}$

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