Question

A U tube with limbs of diameters 5 mm and 2 mm contains water of surface tension \( 7 \times 10^{-2} \, \text{N/m} \), angle of contact is zero and density \( 10^3 \, \text{kg/m}^3 \). The difference in the level in the two limbs is \( (g = 10 \, \text{m/s}^2) \)

• A. 7.7 mm
• B. 6.8 mm
• C. 9.5 mm
• D. 8.4 mm

Hint:

The difference in height in a U-tube with capillary action is inversely proportional to the radius of the tube and is proportional to the surface tension and density.

Answer 1

The Correct Option is D

Step 1: Understanding capillary rise.
The difference in height of the liquid in the two limbs of a U-tube containing a capillary rise is given by the formula: \[ h = \frac{2T \cos \theta}{r \rho g} \] where: - \( T \) is the surface tension, - \( \theta \) is the angle of contact (which is zero here), - \( r \) is the radius of the tube, - \( \rho \) is the density of the liquid, - \( g \) is the acceleration due to gravity. Since the angle of contact is zero, \( \cos \theta = 1 \). The difference in the height of the liquid in the two limbs is: \[ h = \frac{2T}{r \rho g} \] Step 2: Applying the formula for each tube.
For the tube with diameter \( 5 \, \text{mm} \), the radius is \( r_1 = \frac{5}{2} = 2.5 \, \text{mm} = 2.5 \times 10^{-3} \, \text{m} \). For the tube with diameter \( 2 \, \text{mm} \), the radius is \( r_2 = \frac{2}{2} = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \). Now calculate the difference in height for each tube using the surface tension \( T = 7 \times 10^{-2} \, \text{N/m} \), the density \( \rho = 10^3 \, \text{kg/m}^3 \), and \( g = 10 \, \text{m/s}^2 \): For the larger tube: \[ h_1 = \frac{2 \times 7 \times 10^{-2}}{2.5 \times 10^{-3} \times 10^3 \times 10} = 5.6 \, \text{mm} \] For the smaller tube: \[ h_2 = \frac{2 \times 7 \times 10^{-2}}{1 \times 10^{-3} \times 10^3 \times 10} = 14 \, \text{mm} \] Step 3: Total difference in height.
The total difference in the height of the liquid in the two limbs is: \[ h_{\text{total}} = h_2 - h_1 = 14 \, \text{mm} - 5.6 \, \text{mm} = 8.4 \, \text{mm} \] Step 4: Conclusion.
Thus, the difference in the level in the two limbs is 8.4 mm, which corresponds to option (D).