Question

A capillary tube of radius 0.1 mm is dipped in water. The water rises to a height of 2 cm in the tube. If the surface tension of water is 0.072 Nm\(^{-1}\), the contact angle between water and wall of the tube is

• A. \( \cos^{-1} \left( \frac{1}{3.6} \right) \)
• B. \( \cos^{-1} \left( \frac{1}{7.2} \right) \)
• C. \( \cos^{-1} \left( \frac{1}{1.8} \right) \)
• D. \( \cos^{-1} \left( \frac{1}{6.2} \right) \)

Hint:

Capillary rise depends on the radius of the tube, the surface tension, the density of the liquid, and the contact angle.

Answer 1

The Correct Option is B

The capillary rise in a tube is given by the formula: \[ h = \frac{2T \cos \theta}{r \rho g} \] Where: - \( h = 2 \, \text{cm} = 0.02 \, \text{m} \), - \( T = 0.072 \, \text{N/m} \) (surface tension), - \( r = 0.1 \, \text{mm} = 1 \times 10^{-4} \, \text{m} \), - \( \rho \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \)), - \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity), - \( \theta \) is the contact angle. Substitute the known values and solve for \( \cos \theta \): \[ \cos \theta = \frac{r \rho g h}{2T} \] Substituting the values: \[ \cos \theta = \frac{(1 \times 10^{-4}) \times (1000) \times (9.8) \times (0.02)}{2 \times 0.072} \approx \frac{1.96}{0.144} \approx 0.75 \] Thus, the contact angle is: \[ \boxed{\cos^{-1} \left( \frac{1}{7.2} \right)} \]