Question

Three liquids have same surface tension and densities \(ρ_1\), \(ρ_2\) and \(ρ_3\) (\(ρ_1 < ρ_2 < ρ_3\)). In three identical capillaries rise of liquid is same. The corresponding angle of contact \(θ_1\), \(θ_2\) and \(θ_3\) are related as

• A. \(θ_1 < θ_2 < θ_3\)
• B. \(θ_1 > θ_2 > θ_3\)
• C. \(θ_1 = θ_2 = θ_3\)
• D. \(θ_1 > θ_2 < θ_3\)

Hint:

Capillary rise h = 2T cos(theta) / (r rho g), so cos(theta) is proportional to density when h, T, r are fixed.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The rise in a capillary is \(h = \dfrac{2T\cos\theta}{r\rho g}\). Here \(h\), \(T\) and \(r\) are the same for all three liquids.

Step 2: Relate cos theta to density
\[ \cos\theta = \frac{h r g}{2T}\,\rho \Rightarrow \cos\theta \propto \rho \]
Since \(\rho_1 < \rho_2 < \rho_3\), we get \(\cos\theta_1 < \cos\theta_2 < \cos\theta_3\).

Step 3: Convert to angles
For angles between \(0^{\circ}\) and \(90^{\circ}\), cosine decreases as the angle increases. A smaller cosine means a larger angle, so
\[ \theta_1 > \theta_2 > \theta_3 \]
This is option (B).

Final Answer:
The angles satisfy \(\theta_1 > \theta_2 > \theta_3\), option (B). \[ \boxed{\theta_1 > \theta_2 > \theta_3} \]