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(i) Determine the height 'h' to which water rises when capillary tube of radius 0.2 mm is dipped vertically in a beaker containg water (Surface tension of water = 7.0xx10^(-2)Nm^(-1) and angle of contact 0^(@)) (ii) Discuss what will happen if the capillary tube is now pushed down till its height above the surface of water is 4cm. Assume that the density of water is 1000kgm^(-3) and g is 10ms^(-2).

Answer» <html><body><p></p>Solution :i) <a href="https://interviewquestions.tuteehub.com/tag/formula-464310" style="font-weight:bold;" target="_blank" title="Click to know more about FORMULA">FORMULA</a> : `h=(2Scostheta)/(rrhog)` <br/> Angle of contact `theta=0^(@) therefore cos0^(@)=1` <br/> Surface tension `S=7xx10^(-2)Nm^(-1)` <br/> Density of water `d=1000kgm^(-3)` <br/> Acceleration due to gravity `g=10ms^(-2)` <br/> Radius of the tube `r=(0.2//1000)m` <br/> `therefore` Height `h=(<a href="https://interviewquestions.tuteehub.com/tag/2xx-1840186" style="font-weight:bold;" target="_blank" title="Click to know more about 2XX">2XX</a>(7xx10^(-2))xx1)/(((0.2)/(1000))xx(1000)xx(10))=7xx10^(-2)mor7cm`. <br/> <a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) However, when the capillary tube is pushed down till its height above the surface of water is 4 cm, the water rises up to 4 cm only and it will not over flow. In this <a href="https://interviewquestions.tuteehub.com/tag/special-632090" style="font-weight:bold;" target="_blank" title="Click to know more about SPECIAL">SPECIAL</a> case the angle of contact will change. <br/> `(h_(2))/(h_(1))=(costheta_(2))/(costheta_(1))impliescostheta_(2)=(h_(2))/(h_(1))costheta_(1)` <br/> here `h_(1)=7cm, h_(2)=4cm, theta_(1)=0^(@)` <br/> `costheta_(2)=cos^(-1)(0.5714)=55^(@)`</body></html>


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