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. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-10 Mechanical Properties of Fluids



Q.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.

 

 

Answer:-



Radius of the mercury drop, r = 3.00 mm = 3 × 10–3 m
Surface tension of mercury, S = 4.65 × 10–1 N m–1
Atmospheric pressure, P0 = 1.01 × 105 Pa
Total pressure inside the mercury drop
= Excess pressure inside mercury + Atmospheric pressure
= 2S / r + P0
= [ 2 × 4.65 × 10-1 / (3 × 10-3) ] + 1.01 × 105
= 1.0131 × 105
= 1.01 × 105 Pa
Excess pressure = 2S / r
= [ 2 × 4.65 × 10-1 / (3 × 10-3) ] = 310 Pa.