Question

Surface tension of liquid A in a capillary is being measured in the laboratory using capillary rise (refer the figure given below). The capillary radius \( r \) is 100 \(\mu\)m, the height of liquid column \( h \) is 10 cm and \( \Theta = 38^\circ \). Density of air can be neglected. Assume liquid A to have the same density as water. Surface tension of liquid A at room temperature is \(\underline{\hspace{2cm}}\) dyne/cm (round off to one decimal place). 

Hint:

To calculate the surface tension using capillary rise, use the formula \( h = \frac{2 \gamma \cos \Theta}{r \rho g} \) and rearrange to solve for \( \gamma \).

Answer 1

Correct Answer: 61 - 65

We can use the capillary rise formula to calculate the surface tension: \[ h = \frac{2 \gamma \cos \Theta}{r \rho g}, \] where:
- \( h = 10 \, \text{cm} = 0.1 \, \text{m} \) (height of liquid column),
- \( \gamma \) is the surface tension,
- \( \Theta = 38^\circ \) (angle of contact),
- \( r = 100 \, \mu\text{m} = 1 \times 10^{-4} \, \text{cm} \) (capillary radius),
- \( \rho = 1 \, \text{g/cm}^3 \) (density of liquid, same as water),
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).
Rearranging the equation to solve for \( \gamma \): \[ \gamma = \frac{h r \rho g}{2 \cos \Theta}. \] Substitute the known values: \[ \gamma = \frac{0.1 \times 1 \times 10^{-4} \times 1 \times 9.81}{2 \times \cos(38^\circ)}. \] Using \( \cos(38^\circ) \approx 0.788 \): \[ \gamma = \frac{0.1 \times 1 \times 10^{-4} \times 9.81}{2 \times 0.788} \approx 0.0615 \, \text{dyne/cm}. \] Thus, the surface tension of liquid A is approximately \( 0.6 \, \text{dyne/cm} \).