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

Step 1: Understanding the Concept:
Due to surface tension, liquid rises in a capillary tube. In a U-tube with different bore radii, the height of the liquid rise will be different in each limb, creating a level difference.
Step 2: Key Formula or Approach:
Capillary rise \(h\) is given by:
\[ h = \frac{2T \cos \theta}{r \rho g} \]
Difference in level \(\Delta h = h_1 - h_2\).
Step 3: Detailed Explanation:
Given: \(T = 7.3 \times 10^{-2} \text{ N/m}\), \(\rho = 10^3 \text{ kg/m}^3\), \(g = 10 \text{ m/s}^2\), \(\cos(0^\circ) = 1\).
Radius of limb 1 (\(r_1\)) = 2.5 mm = \(2.5 \times 10^{-3} \text{ m}\).
Radius of limb 2 (\(r_2\)) = 4.0 mm = \(4.0 \times 10^{-3} \text{ m}\).
\[ h_1 = \frac{2 \times 7.3 \times 10^{-2}}{2.5 \times 10^{-3} \times 10^3 \times 10} = \frac{14.6 \times 10^{-2}}{25} = 0.584 \times 10^{-2} \text{ m} = 5.84 \text{ mm} \]
\[ h_2 = \frac{2 \times 7.3 \times 10^{-2}}{4.0 \times 10^{-3} \times 10^3 \times 10} = \frac{14.6 \times 10^{-2}}{40} = 0.365 \times 10^{-2} \text{ m} = 3.65 \text{ mm} \]
Difference \(\Delta h = 5.84 - 3.65 = 2.19 \text{ mm}\).
Step 4: Final Answer:
The difference in level is 2.19 mm.