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 experiments is 7.30 xx 10^(-2) "N/m". 1 atmospheric pressure =1.01 xx 10^(5) Pa density of water =7000" kg/m"^(3), g="9.80 ms"^(-2). Also calculate the excess pressure.

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 water at the temperature of the experiments is 7.30 xx 10^(-2) "N/m". 1 atmospheric pressure =1.01 xx 10^(5) Pa density of water =7000" kg/m"^(3), g="9.80 ms"^(-2). Also calculate the excess pressure.
Answer» Solution :Given, `R=(2.00 mm)/(2)=1.00 mm` `=1.00 xx 10^(-3)m` `h=8.00 cm =8 xx 10^(-2) m` `sigma=7.30 xx 10^(-2) N//m` `P_("atm")=1.01 xx 10^(5) Pa` `p=1000 kg//m^(3), g=9.80 m//s^(2)` EXCESS pressure inside the bubble is `p=(2sigma)/(r)=(2 xx 7.30 xx 10^(-2))/(1 xx 10^(-3)) =146 Pa` Pressure required in the TUBE=Atmospheric pressure +Pressure due to water column `=P_("atm")+HPG` `=1.01 xx 10^(5)+8 xx 10^(-2) xx 10^(-3) xx 9.80` `=1.01 xx 10^(5) +0.00784 xx 10^(5)` `=1.01784 xx 10^(5) Pa` Pressure inside the bubble =Pressure outside the tube +Exces pressure `=1.01784 xx 10^(5)+146` `=1.0193 xx 10^(5) Pa or` `=1.02 xx 10^(5) Pa`