The lower end of a capillary tube ofdiameter 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 the end in water ?The surface tension of water at temperature of the experiments is 7.30xx10^(-2)Nm^(9-1). 1 atmospheric pressure =1.01xx10^(5)Pa.Density of water =1000kg//m^(3),g=9.80ms^(-2).Also calulate the excess pressure .

The lower end of a capillary tube ofdiameter 2.00 mm is dipped 8.00 cm

1.	The lower end of a capillary tube ofdiameter 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 the end in water ?The surface tension of water at temperature of the experiments is 7.30xx10^(-2)Nm^(9-1). 1 atmospheric pressure =1.01xx10^(5)Pa.Density of water =1000kg//m^(3),g=9.80ms^(-2).Also calulate the excess pressure .
Answer» Solution :Radius of capillary tube , `r=(2.0)/(2)mm` `=1.0mm=1XX10^(-3)m` Surface tension of water `=7.30xx10^(-2)Nm^(-1)` Atmospheric PRESSURE `P=1.01xx10^(5)Pa` Density of water `rho=1000kgm^(-3)` Acceleration of gravity `g=9.8ms^(-2)` DEPTH of bubble `h=8.0cm=8xx10^(-2)m` The pressure difference between inside and outside the bubble form in liquid, `P_(i)-P_(o)=(2s)/(r)` The radius if bubble is equal to the radius of capillary tube) `thereforeP_(i)=P_(o)+(2s)/(r)` ...(1) But `P_(o)=P_(a)+hrhog` `=1.01xx10^(5)+8xx10^(-2)xx10^(3)xx9.8` `=1.01xx10^(5)+784` `1.01784xx10^(5)Nm^(-2)` From equation (1), `P_(i)=P_(o)+(2s)/(r)` `=1.01784xx10^(5)+(2xx7.3xx10^(-2))/(1xx10^(-3))` `=1.01784xx10^(5)+14.6xx10` `=1.01784+146` `=1.01930` `=1.02xx10^(5)N//m^(2)` The excess pressure in a bubble , `P_(i)-P_(o)=1.01930xx10^(5)-1.01784xx10^(5)` `=0.00146xx10^(5)` `=146Pa`