Question

Two capillary tubes of radii \(r_1\) and \(r_2\) \((r_1 > r_2)\) are dipped vertically in the same liquid. The rise of liquid in the tubes \(h_1\) and \(h_2\) satisfies:

• A. \(h_1 > h_2\)
• B. \(h_1 < h_2\)
• C. \(h_1 = h_2\)
• D. \(h_1 r_2 = h_2 r_1\)

Hint:

Remember: "Thinner tubes draw higher columns." The narrower the capillary bore, the higher the liquid column must ascend to balance the upward surface tension force against gravity.

Answer 1

The Correct Option is B

Concept: The height \(h\) to which a liquid rises (or falls) in a capillary tube is governed by Jurin's Law. For a liquid with surface tension \(T\), density \(\rho\), and contact angle \(\theta\) inside a tube of radius \(r\) under gravity \(g\), the equilibrium height is given by the formula: \[ h = \frac{2T \cos\theta}{r \rho g} \] Since the tubes are dipped in the same liquid, the physical parameters \(T\), \(\rho\), \(\theta\), as well as the gravitational constant \(g\), are exactly identical for both cases. This implies that the capillary rise is inversely proportional to the tube radius: \[ h \propto \frac{1}{r} \quad \Rightarrow \quad h \cdot r = \text{constant} \] Step 1: Set up the proportionality relationship.
Using the inverse relationship from Jurin's Law, we can relate the heights and radii of the two tubes: \[ h_1 r_1 = h_2 r_2 \]

Step 2: Analyze the given inequality configuration.
We are given that the radius of the first tube is larger than the second tube: \[ r_1 > r_2 \] Since height is inversely proportional to the radius (\(h \propto \frac{1}{r}\)), a larger radius results in a smaller capillary rise. Therefore: \[ h_1 < h_2 \] This directly corresponds to option (B).