The lower end of a capillary tube of diameter $2.00 \; mm$ is dipped $8.00 \; cm$ below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water? The surface tension of water at the temperature of the experiment is $7.30 \times 10^{-2} \; N m^{-1}$. Atmospheric pressure $= 1.01 \times 10^{5} \; Pa$,density of water $= 1000 \; kg m^{-3}$,$g = 9.80 \; m s^{-2}$. Also,calculate the excess pressure.

(N/A) The excess pressure in a bubble of gas in a liquid is given by $P_{ex} = 2S/r$,where $S$ is the surface tension of the liquid-gas interface.
Here,the diameter of the capillary tube is $d = 2.00 \; mm = 2.00 \times 10^{-3} \; m$,so the radius of the hemispherical bubble is $r = d/2 = 1.00 \times 10^{-3} \; m$.
The excess pressure is $P_{ex} = 2S/r = (2 \times 7.30 \times 10^{-2} \; N m^{-1}) / (1.00 \times 10^{-3} \; m) = 146 \; Pa$.
The pressure outside the bubble at a depth $h = 8.00 \; cm = 0.08 \; m$ is $P_o = P_{atm} + h \rho g$.
$P_o = 1.01 \times 10^5 \; Pa + (0.08 \; m \times 1000 \; kg m^{-3} \times 9.80 \; m s^{-2}) = 1.01 \times 10^5 \; Pa + 784 \; Pa = 101784 \; Pa$.
The total pressure required inside the tube is $P_i = P_o + P_{ex} = 101784 \; Pa + 146 \; Pa = 101930 \; Pa = 1.0193 \times 10^5 \; Pa$.