One end of a uniform glass capillary tube of radius $r = 0.025\ cm$ is immersed vertically in water to a depth $h = 1\ cm$. The excess pressure in ${Nm}^{- 2}$ required to blow an air bubble out of the tube (Surface tension of water = $7 imes 10^{- 2}\ {Nm}^{- 1}$, Density of water = $10^{3}\ kg\ m^{- 3}$ , Acceleration due to gravity= 10${ms}^{- 2}$)

Question

One end of a uniform glass capillary tube of radius $r = 0.025\ cm$ is immersed vertically in water to a depth $h = 1\ cm$. The excess pressure in ${Nm}^{- 2}$ required to blow an air bubble out of the tube (Surface tension of water = $7 	imes 10^{- 2}\ {Nm}^{- 1}$, Density of water = $10^{3}\ kg\ m^{- 3}$ , Acceleration due to gravity= 10${ms}^{- 2}$)

Dr. Abubacker · Accepted Answer

(b) Excess pressure
$p = hdg + h^{'}dg$
$\Longrightarrow \ p = dg\left( h + h^{'} ight)$
$Where\ h\ is\ capillary\ rise = \frac{2T}{rdg}$
$= \frac{2 	imes 7 	imes 10^{- 2}}{25 	imes 10^{- 5} 	imes 10^{3} 	imes 10} = 0.056\ m$
$	herefore p = 10^{3} 	imes 10\lbrack 0.056 + 0.01brack$
$= 0.066 	imes 10^{4}$
$= 0.0066 	imes 10^{5}\ {Nm}^{- 2}$

Explanation

Reviewed by Dr. Abubacker

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