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 waterat temperature of the experiment is 7.30xx10^(-2)Nm^(-1). 1 atmospheric pressure = 1.01xx10^(5) Pa, density of water = 1000 kg m^(-3),g=9.80ms^(-2)].

The lower end of a capillary tube of diameter 2.00 mm is dipped 8.00 c

1.	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 waterat temperature of the experiment is 7.30xx10^(-2)Nm^(-1). 1 atmospheric pressure = 1.01xx10^(5) Pa, density of water = 1000 kg m^(-3),g=9.80ms^(-2)].
Answer» Solution :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 thiscase. The radius of the bubble is r. Now the pressure OUTSIDE the bubble, `p_("out")` 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 inside the bubble is `P_("in")=P_("out")+2S//r=1.01784xx10^(5)+(2xx7.3xx10^(-2)//10^(-3))` `=(1.01784+0.00146)xx10^(5)=1.02xx10^(5)Pa`