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).
Given that r1=23.0=1.5mm=1.5×10−3m,r2=26.0=3.0mm=3.0×10−3m T=7.3×10−2Nm−1,θ=0∘ρ=1.0×103kgm−1,g=9.8ms−2 When angle of contact is zero degree the radius of the meniscus equals radius of bore Excess pressure in the first bore P1=r22T=1.5×10−32×7.3×10−2=97.3 Pascal Excess pressure in the second bore,P2=r22T=3×10−32×7.3×10−2=48.7 Pascal Hence, pressure difference in the two limbs of the tube ΔP=P1−P2=hρg or h=ρgP1−P2=1.0×103×9.897.3−48.7=5.0mm
