Find the excess pressure inside (a) a drop of mercury of radius 2 mm (b) a soap bubble of radius 4 mm and

 Excess pressure inside a drop is given as p = 2S/r. 

(a) The radius of mercury drop r = 2 mm =0.002 m 

The surface tension of mercury Sₘ =0.465 N/m 

Excess pressure inside = 2Sₘ/r =2*0.465/0.002 N/m² 

= 465 N/m²  

(b) The radius of the soap bubble = r'=4 mm =0.004 m 

The surface tension of the soap surface Sₓ =0.03  N/m 

Excess pressure inside a soap bubble = 4Sₓ/r' 

{Since it has two liquid surfaces} 

=4*0.03/0.004 N/m² = 30 N/m²  

(c) The surface tension of water, Sᵥ =0.076 N/m 

The radius of the water bubble r = 4 mm =0.004 m 

Excess pressure inside the water bubble in the tank = 2Sᵥ/r 

=2*0.076/0.004 N/m² = 38 N/m²