Question

What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? The 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.

Answer 1

Radius of the mercury drop,

r = 3mm = 3 × 10^-3m Surface tension of mercury, S = 4.65 × 10^-1 Nm^-1. 

Atmospheric pressure, P_o = 1.01 × 10⁵ Pa

Total pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure

= 2S/r + p_o

  + 1.01 × 10⁵

Total Pressure = 1.013 × 10⁵ Pa

Excess pressure = 2S/r

= 310 Pa.