A glass capillary tube of inner diameter 0.28 | vedclass

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(D) The capillary rise $h$ is given by $h = \frac{2T \cos 	heta}{r ho g}$.To bring the water level in the capillary to the same level as the vessel

Vedclass · Accepted Answer

$101 	imes 10^3$

Vedclass · Answer

(D) The capillary rise $h$ is given by $h = \frac{2T \cos 	heta}{r ho g}$.To bring the water level in the capillary to the same level as the vessel,we must apply an excess pressure $P$ equal to the capillary pressure $h ho g$.Thus,$P = h ho g = \frac{2T \cos 	heta}{r}$.Given: $T = 0.07 \ N/m$,$d = 0.28 \ mm = 0.28 	imes 10^{-3} \ m$,so $r = 0.14 	imes 10^{-3} \ m$,and for water $	heta = 0^{\circ}$ (so $\cos 	heta = 1$).$P = \frac{2 	imes 0.07}{0.14 	imes 10^{-3}} = \frac{0.14}{0.14 	imes 10^{-3}} = 10^3 \ N/m^2$.This is the excess pressure required. The total pressure to be applied on the water surface in the capillary tube is the sum of the atmospheric pressure and this excess pressure.Total Pressure $= P_{atm} + P = 10^5 + 10^3 = 100 	imes 10^3 + 1 	imes 10^3 = 101 	imes 10^3 \ N/m^2$.

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