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

Step 1: Understanding the Concept:
The height $h$ to which a liquid rises in a capillary tube of radius $r$ is given by Jurin's Law. It depends on the surface tension, the contact angle, and the tube radius.

Step 2: Key Formula or Approach:
The formula for capillary rise is:
\[ h = \frac{2T \cos \theta}{\rho g r} \]
Where $T$ is surface tension, $\rho$ is density, and $r$ is radius.

Step 3: Detailed Explanation:
For the same liquid (or liquids with the same surface tension \( T_1 = T_2 \)) and same density \( \rho \), the product \( h \cdot r \) is constant:
\[ h \cdot r = \frac{2T \cos \theta}{\rho g} = \text{constant} \]
Given that \( r_1>r_2 \), it follows that:
\[ h_1 \cdot r_1 = h_2 \cdot r_2 \]
Since \( r_1>r_2 \), we must have \( h_1<h_2 \) to satisfy the equality.
This physically corresponds to the fact that narrower tubes result in a higher capillary rise for the same surface tension.

Step 4: Final Answer:
The correct physical relation is \( h_1<h_2 \) and \( T_1 = T_2 \).