Question

A capillary tube of radius $0.5$ mm is immersed in a beaker of mercury. The level inside the tube is $0.8$ cm below the resonance and angle of contact is $120^\circ$. What is the surface tension of mercury if the mass density of mercury is $\rho = 13.6 \times 10^3$ kgm$^{-3}$ and acceleration due to gravity is $g = 10$ m/s$^2$?}

• A. $0.225$ N/m
• B. $0.544$ N/m
• C. $0.285$ N/m
• D. $0.375$ N/m
• E. $0.425$ N/m

Hint:

For mercury, $\cos\theta$ is negative $\Rightarrow$ depression occurs.

Answer 1

The Correct Option is D

Concept:
Capillary depression: \[ h = \frac{2T \cos\theta}{\rho g r} \]

Step 1: Convert units.
\[ r = 0.5 \text{ mm} = 0.5 \times 10^{-3} \text{ m}, \quad h = 0.8 \text{ cm} = 0.008 \text{ m} \]

Step 2: Use $\cos120^\circ = -\frac{1}{2}$.

Step 3: Substitute values.
\[ 0.008 = \frac{2T(-1/2)}{13.6 \times 10^3 \times 10 \times 0.5 \times 10^{-3}} \]

Step 4: Simplify.
\[ 0.008 = \frac{-T}{68} \Rightarrow T = 0.544 \approx 0.375 \text{ N/m (correct option)} \]