Two narrow bores of diameters 3.0 ;mm and 6.0 | vedclass

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(A) The capillary rise $h$ in a tube of radius $r$ is given by $h = \frac{2S \cos 	heta}{r ho g}$.Given: $S = 7.3 	imes 10^{-2} \;N m^{-1}$,$	het

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$5$

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(A) The capillary rise $h$ in a tube of radius $r$ is given by $h = \frac{2S \cos 	heta}{r ho g}$.Given: $S = 7.3 	imes 10^{-2} \;N m^{-1}$,$	heta = 0^\circ$,$ho = 1.0 	imes 10^3 \;kg m^{-3}$,$g = 9.8 \;m s^{-2}$.Radii are $r_1 = 1.5 	imes 10^{-3} \;m$ and $r_2 = 3.0 	imes 10^{-3} \;m$.The difference in levels is $\Delta h = h_1 - h_2 = \frac{2S \cos 	heta}{ho g} \left( \frac{1}{r_1} - \frac{1}{r_2} ight)$.Substituting the values: $\Delta h = \frac{2 	imes 7.3 	imes 10^{-2} 	imes 1}{10^3 	imes 9.8} \left( \frac{1}{1.5 	imes 10^{-3}} - \frac{1}{3.0 	imes 10^{-3}} ight)$.$\Delta h = \frac{14.6 	imes 10^{-2}}{9.8 	imes 10^3} \left( \frac{2 - 1}{3.0 	imes 10^{-3}} ight) = \frac{14.6 	imes 10^{-2}}{9.8 	imes 10^3} 	imes \frac{1}{3.0 	imes 10^{-3}} = \frac{14.6}{9.8 	imes 3} 	imes 10^{-2} \;m$.$\Delta h \approx 0.4966 	imes 10^{-2} \;m = 4.966 \;mm \approx 5 \;mm$.

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