Question

Surface tension of two liquids (having same densities), $T_1$ and $T_2$ are measured using capillary rise method utilizing two tubes with inner radii of $r_1$ and $r_2$ where $r_1>r_2$. The measured liquid heights in these tubes are $h_1$ and $h_2$ respectively. [Ignore the weight of the liquid about the lowest point of meniscus]. The heights $h_1$ and $h_2$ and surface tensions $T_1$ and $T_2$ satisfy the relation :

• A. $h_1>h_2$ and $T_1<T_2$
• B. $h_1 = h_2$ and $T_1 = T_2$
• C. $h_1<h_2$ and $T_1 = T_2$
• D. $h_1>h_2$ and $T_1 = T_2$

Hint:

Remember Jurin's Law: $h \propto 1/r$. For the same liquid, height is inversely proportional to the radius of the tube.

Answer 1

The Correct Option is C

To solve this problem, we need to understand the relationship between surface tension, capillary rise, and the radius of the capillary tube used in the measurement. The capillary rise method is based on the formula for capillary rise, which is given as:

\(h = \frac{2T}{\rho g r}\) 

Where:

• \(h\) = height of the liquid column
• \(T\) = surface tension of the liquid
• \(\rho\) = density of the liquid
• \(g\) = acceleration due to gravity
• \(r\) = radius of the capillary tube

Given that the two liquids have the same density, the formula indicates that for a fixed \(\rho\) and \(g\), the height \(h\) is directly proportional to \(T\) and inversely proportional to \(r\).

For liquid 1 in the tube with radius \(r_1\) and liquid 2 in the tube with radius \(r_2\), the heights \(h_1\) and \(h_2\) respectively can be expressed as:

\(h_1 = \frac{2T_1}{\rho g r_1}\)

\(h_2 = \frac{2T_2}{\rho g r_2}\)

Given \(r_1 > r_2\), it follows from the inverse proportionality to \(r\) that \(h_1 < h_2\) if the surface tensions are equivalent, i.e., \(T_1 = T_2\).

This leads us to conclude that the correct option is \(h_1<h_2\) and \(T_1 = T_2\), satisfying the relationship given by the capillary rise method. Hence, the correct answer to the question is:

Option: $h_1<h_2$ and $T_1 = T_2$