Question

Two narrow bores of diameter 5.0 mm and 8.0 mm are joined together to form a U-shaped tube open at both ends. If this U-tube contains water, what is the difference in the level of two limbs of the tube. [Take surface tension of water $T = 7.3 \times 10^{-2}$ $Nm^{-1}$, angle of contact $= 0$, $g = 10$ $ms^{-2}$ and density of water $= 1.0 \times 10^3$ kg $m^{-3}$]

• A. 5.34 mm
• B. 3.62 mm
• C. 4.97 mm
• D. 2.19 mm

Hint:

Remember: \(h \propto 1/r\). The narrower the tube, the higher the liquid will rise. Be careful not to use diameter instead of radius in the formula.

Answer 1

The Correct Option is D

To solve this problem, we need to determine the difference in water levels in the two limbs of a U-shaped tube with different diameters due to the surface tension of water. The given parameters are surface tension \( T = 7.3 \times 10^{-2} \, \text{Nm}^{-1} \), angle of contact \( \theta = 0 \) degrees, gravitational acceleration \( g = 10 \, \text{ms}^{-2} \), and density of water \( \rho = 1.0 \times 10^3 \, \text{kg/m}^{3} \).

The height difference due to capillarity in a tube is given by the formula: 

\(h = \frac{2T \cos \theta}{\rho g r}\)

where \( r \) is the radius of the bore of the tube.

1. First, calculate the radius of each tube. The diameters given are 5.0 mm and 8.0 mm. Thus:
   • Radius of the first bore \( r_1 = \frac{5.0 \, \text{mm}}{2} = 2.5 \, \text{mm} = 2.5 \times 10^{-3} \, \text{m} \).
   • Radius of the second bore \( r_2 = \frac{8.0 \, \text{mm}}{2} = 4.0 \, \text{mm} = 4.0 \times 10^{-3} \, \text{m} \).
2. Calculate the capillary rise in each limb using the formula:
   • \(h_1 = \frac{2 \times 7.3 \times 10^{-2} \times \cos 0}{1.0 \times 10^3 \times 10 \times 2.5 \times 10^{-3}} = \frac{14.6 \times 10^{-2}}{1.0 \times 25 \times 10^{-3}} = \frac{14.6}{25} = 0.584 \, \text{m} = 584 \, \text{mm}\)
   • \(h_2 = \frac{2 \times 7.3 \times 10^{-2} \times \cos 0}{1.0 \times 10^3 \times 10 \times 4.0 \times 10^{-3}} = \frac{14.6 \times 10^{-2}}{1.0 \times 40 \times 10^{-3}} = \frac{14.6}{40} = 0.365 \, \text{m} = 365 \, \text{mm}\)
3. Determine the difference in water levels:
   • \(\Delta h = h_1 - h_2 = 584 \, \text{mm} - 365 \, \text{mm} = 219 \, \text{mm} = 2.19 \, \text{mm}\)

Therefore, the difference in the level of two limbs of the tube is 2.19 mm. This is why the correct answer is 2.19 mm.