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Q 10.30 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).

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For the angle of contact \theta, radius of the tube r, surface tension t, the density of fluid \rho the rise in the column is given by

h=\frac{2Tcos\theta }{r\rho g}

The radii of the two limbs r1 and r2 are 3.0 mm and 1.5 mm respectively 

The level in the limb of diameter 6.0 mm is

\\h_{1}=\frac{2Tcos\theta }{r_{1}\rho g}\\ h_{1}=\frac{2\times 7.3\times 10^{-2}cos0^{o}}{3\times 10^{-3}\times 10^{3}\times 9.8}\\h_{1}=4.97\times 10^{-3}m

The level in the limb of diameter 3.0 mm is

\\h_{2}=\frac{2Tcos\theta }{r_{2}\rho g}\\ h_{2}=\frac{2\times 7.3\times 10^{-2}cos0^{o}}{1.5\times 10^{-3}\times 10^{3}\times 9.8}\\h_{2}=9.93\times 10^{-3}m

The difference in the heights is h2 - h1 = 4.96 mm

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