The lower end of a capillary tube of diameter 2.00mm is dipped 8.00cm 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 `7.30 times 10^2 Nm^-1`.1 atmospheric pressure `=1.01 times 10^5 Pa` density of water `=1000 kg//m^3,g=9.80 ms^-2`. Also calculate the excess pressure.

Explanation

The excess pressure in a bubble of gas in a liquid is given by `2sigma//r`, where `sigma` 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 `4sigma//r`]. The radius of the bubble is r. Now the pressure outside the bubble, pout equals atmospheric pressure plus the pressure due to 8.00cm of water column.
`p_("out")=(1.01xx10^(5)+0.08xx1000xx9.80)Pa=1.01784xx10^(5)Pa`
Therefore, the pressure in side the bubble is `p_("in")=p_("out")+2sigma//r`
= `1.01784xx10^(5)+(2xx7.3xx10^(-2)//10^(-3))=(1.01784+0.00146)xx10^(5)=1.02xx10^(5)Pa`