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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» <html><body><p></p>Solution :Radius of capillary tube ,<br/>`r=(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.0)/(2)mm`<br/>`=1.0mm=<a href="https://interviewquestions.tuteehub.com/tag/1xx10-1804645" style="font-weight:bold;" target="_blank" title="Click to know more about 1XX10">1XX10</a>^(-3)m`<br/>Surface tension of water `=7.30xx10^(-2)Nm^(-1)`<br/>Atmospheric <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> `P=1.01xx10^(5)Pa`<br/>Density of water `rho=1000kgm^(-3)`<br/>Acceleration of gravity `g=9.8ms^(-2)`<br/><a href="https://interviewquestions.tuteehub.com/tag/depth-948801" style="font-weight:bold;" target="_blank" title="Click to know more about DEPTH">DEPTH</a> of bubble `h=8.0cm=8xx10^(-2)m`<br/>The pressure difference between inside and outside the bubble form in liquid, <br/> `P_(i)-P_(o)=(2s)/(r)`<br/>The radius if bubble is equal to the radius of capillary tube)<br/>`thereforeP_(i)=P_(o)+(2s)/(r)` ...(1)<br/>But `P_(o)=P_(a)+hrhog`<br/>`=1.01xx10^(5)+8xx10^(-2)xx10^(3)xx9.8`<br/>`=1.01xx10^(5)+784`<br/>`1.01784xx10^(5)Nm^(-2)`<br/>From equation (1),<br/>`P_(i)=P_(o)+(2s)/(r)`<br/>`=1.01784xx10^(5)+(2xx7.3xx10^(-2))/(1xx10^(-3))`<br/>`=1.01784xx10^(5)+14.6xx10`<br/>`=1.01784+146`<br/>`=1.01930`<br/>`=1.02xx10^(5)N//m^(2)`<br/>The excess pressure in a bubble , `P_(i)-P_(o)=1.01930xx10^(5)-1.01784xx10^(5)`<br/>`=0.00146xx10^(5)`<br/>`=146Pa`</body></html>


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