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 × 10⁻² Nm⁻¹, angle of contact = 0, g = 10 ms² and density of water = 1.0 × 10³ kg m⁻³]

Solution

Physics sir · 4 steps

Each limb has its own capillary rise, and the level difference is just the narrow rise minus the wide rise.

1. Why levels differ. Just under a curved meniscus the pressure is lower than atmospheric by (2T cos θ)/r. The narrower limb has the larger pressure drop, so water stands higher there.
2. Rise in each limb. With θ = 0, each limb would rise by h = 2T/(rρ g). The radii are r₁ = 2.5 mm and r₂ = 4.0 mm, half of the given diameters.

   Δ h = h₁ - h₂ = 2T/(ρ g)(1/r₁ - 1/r₂)

3. Put in numbers. 2T/(ρ g) = (2 × 7.3 × 10⁻²)/(10³ × 10) = 1.46 × 10⁻⁵ m². The bracket is 1/(2.5 × 10⁻³) - 1/(4.0 × 10⁻³) = 400 - 250 = 150 m⁻¹.
4. So. Δ h = 1.46 × 10⁻⁵ × 150 = 2.19 × 10⁻³ m = 2.19 mm. The narrow limb is higher. The answer is (b).

The idea to keep. In a U-tube with unequal bores, find each capillary rise h = (2T cos θ)/(rρ g) and subtract. Always halve the diameter to get the radius first.

Answer from the NTA final answer key. Solution by PrepWiser.