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  • Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the

Q 10.29 Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is 0.465 N m-1. Density of mercury = 13.6 × 103 kg m-3.

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Since the angle of contact is obtuse the Pressure will be more on the Mercury side.

This pressure difference is given as

\\\Delta P=\frac{2Tcos(180-\theta )}{r}\\ \Delta P=\frac{2\times 0.465cos40^{o}}{10^{-3}}\\\Delta P=\frac{2\times 0.465\times 0.766}{10^{-3}}\\ \Delta P=712.38\ Pa

The dip of mercury inside the narrow tube would be equal to this pressure difference

\\\Delta P=\rho _{Hg}hg\\ h=\frac{\Delta P}{\rho _{Hg}g}\\ h=\frac{712.38}{13.6\times 10^{3}\times 9.8}\\ h=5.34\times 10^{-3}m

The mercury dips down in the tube relative to the liquid surface outside by an amount of 5.34 mm. 

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