Question

A glass tube of inside diameter 1.0 mm is dipped vertically into a container of mercury, with its lower end 1 cm below the mercury surface.

a) What must be the gauge pressure of the air in the tube to blow a hemispherical bubble at its lower end? (Answer: 1400 Pa)

b) To what height will mercury rise in the tube if the air pressure in the tube is 3000 Pa below atmospheric? The contact angle is 140 degrees and the density of mercury is 13.6 g/cm^3. (Answer: 2.3 cm)

Answer 1

(a) In this case we apply bernoullis equation at the surface and below 1 cm in the mercury.
     

      At surface the pressure will be atmospheric so put it equal to zero , kinetic energy will be zero at both section.     Hence we get

on putting the value we get P₂ = 1400 Pa
(b) Simply putting P= (density)gh     (P= 3000, g = 9.81 , )
we will get the value of h =2.25cm which on roundoff will give h= 2.3 cm