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.
Radius of the mercury drop, r=3.00mm=3×10−3 m Surface tension of mercury, S=4.65×10−1 N/m Atmospheric pressure, Po=1.01×105 Pa Total pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure = 2S/r+Po = [2×4.65×10−1/(3×10−3)]+1.01×105 =1.0131×105 Excess pressure = 2S / r = [2×4.65×10−1/(3×10−3)]=310 Pa
