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InterviewSolution
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A glass tube of 1 mm bore is dipped vertically into a container of mercury, with its lower end 5 cm below the mercury surface. What must be the gauge pressure of air in the tube to a hemispherical bubble at its lower end? Given density of mercury ` = 13.6 xx 10^(3) kg//m^(3)`, surface tension of mercury = `440 xx 10^(-3) Nm^(-1)` and `g = 10m//s^(2)` |
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Answer» Here, `R=1.2 mm = 0.5 mm` `= 5 xx 10^(-4)m`, `h= 5 cm = 0.05 m , rho = 13.6 xx 10^(3) kg//m^(3)` `S= 440 xx 10^(-3) Nm^(-1)` Gauge pressure in hemispherical bubble `= h rho g + (2S)/(R )` `= 0.05 xx (13.6 xx 10^(3)) xx 10 + ( 2 xx 440 xx10^(3))/(5 xx 10^(-4))` `= 6800+1760 = 8560 Pa`. |
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