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  • What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature ? Surface tension of mercury at that temperature (20 °C) is 4.65 × 10^–1 N m^–1. The atmospheric pressure is

Q 10.19 What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20 °C)  is 4.65 × 10-1 N m-1. The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop.

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Surface Tension of Mercury is T=4.65\times 10^{-1}\ N\ m^{-1}

The radius of the drop of Mercury is r = 3.00 mm

Excess pressure inside the Mercury drop is given by

\\\Delta P=\frac{2T}{r}\\ \Delta P=\frac{2\times 4.65\times 10^{-1}}{3\times 10^{-3}}\\ \Delta P=310\ Pa

Atmospheric Pressure is P_{0}=1.01\times 10^{5}\ Pa

Total Pressure inside the Mercury drop is given by

\\P_{T}=\Delta P + P_{0}\\ P_{T}=310+1.01\times 10^{5}\\ P_{T}=1.0131\times 10^{5}

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