A glass capillary tube of inner diameter 0.28 | vedclass

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(D) Given:Inner diameter $d = 0.28 \ mm = 0.28 	imes 10^{-3} \ m$.Radius $r = d/2 = 0.14 	imes 10^{-3} \ m$.Surface tension $T = 0.07 \ N/m$.For wat

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$101 	imes 10^3$

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(D) Given:Inner diameter $d = 0.28 \ mm = 0.28 	imes 10^{-3} \ m$.Radius $r = d/2 = 0.14 	imes 10^{-3} \ m$.Surface tension $T = 0.07 \ N/m$.For water,the contact angle $	heta = 0^\circ$,so $\cos 	heta = 1$.Due to surface tension,the capillary rise $h$ is given by $h = \frac{2T}{r ho g}$.The pressure difference across the meniscus is $\Delta P = \frac{2T}{r}$.To keep the water level in the tube the same as in the vessel,we must apply an external pressure equal to the capillary pressure difference.$\Delta P = \frac{2 	imes 0.07}{0.14 	imes 10^{-3}} = \frac{0.14}{0.14 	imes 10^{-3}} = 10^3 \ N/m^2$.Since the water surface in the tube is depressed by this pressure relative to the atmospheric pressure,the total pressure required to maintain the level is the sum of atmospheric pressure and the capillary pressure:$P_{total} = P_{atm} + \Delta P = 10^5 + 10^3 = 100 	imes 10^3 + 1 	imes 10^3 = 101 	imes 10^3 \ N/m^2$.

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