Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 × 10–2 N m–1. Take the angle of contact to be zero and density of water to be 1.0 × 103 kg m–3 (g = 9.8 m s–2).
Diameter of the first tube = 3 mm
Diameter of the second tube = 6 mm
Thus, Radius of the first tube, r1 = 1.5 mm = 1.5 × 10-3 m
Radius of the first tube, r2 = 3 mm = 3 × 10-3 m
Surface tension of water, S = 7.3 × 10-2 N/m
Density of water, ρ = 1000 kg/m3
Angle of contact between bore surface and water, θ = 00
Acceleration due to gravity, g = 9.8 m/s
Height of water rise, 
Where,
S = Surface tension
θ = Contact angle
ρ = Density
g = acceleration due to gravity
r = radius
Let h1 and h2 be the height rise of water in tube1 and tube2 respectively. Then the height are given by,
; 
Thus, the difference between heights is given by,
Δh = h1 – h2
= 4.996 × 10-3 m
≈ 4.97 mm
∴ The difference between the water levels in two bores = 4.97 mm