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 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 .

Explanation

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`