The lower end of 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 temperature of the experiments is `7.30xx10^(-2)Nm^(1)`. `1` atmospheric pressure `=1.01xx10^(5)` Pa, density of water `=1000 "kg"//"m"^(3)`, `g=9.8xxms^(-2)`. Also calculate the excess pressure .

Explanation

The excess pressure in a bubble of gas in a liquid is given by `2S//r`, where S is the surface tension of the liquid gas interface. You should note there is only one liquid surface in this case. (For a bubble of liquid in a gas, there are two liquid surfaces, so the formula for excess pressure in that case is `4S//r`.) The radius of the bubble is r. Now the pressure outside the bubble, within water `P_(0)` equals atmospheric pressure plus the pressure due to `8.00` cm of water column. That is
`P_(0)=(1.01xx10^(5)Pa+0.08mxx1000kgm^(-3)xx9.80ms^(-1))`
`=1.01784xx10^(5)` Pa
Therefore, the pressure inside the bubble is `P_(1)=P_(0)+2S//r(as r=1mm)`
`=1.01784xx10^(5)Pa+(2xx7.3xx10^(-2)Pam//10^(-3)m)`
`=(1.01784+0.00146)xx10^(5)Pa`
`=1.02xx10^(5)Pa`
where the radius of the bubble is taken to be equal to the radius of the capillary tube, since the bubble is hemispherical `!` (The answer has been rounded off to three significant figures). The excess pressure in the bubble is `146` Pa from `(1.0178+0.00146)xx10^(5)` Pa.