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(B) The vertical height $h$ of the water column in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{r ho g}$.Here,$T = 7 	imes

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$8$

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(B) The vertical height $h$ of the water column in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{r ho g}$.Here,$T = 7 	imes 10^{-2} \ N/m$,$r = 0.35 \ mm = 0.35 	imes 10^{-3} \ m$,$	heta = 0^{\circ}$ (for glass-water contact),$ho = 10^3 \ kg/m^3$,and $g = 10 \ m/s^2$.Substituting the values:$h = \frac{2 	imes (7 	imes 10^{-2}) 	imes \cos 0^{\circ}}{(0.35 	imes 10^{-3}) 	imes 10^3 	imes 10} = \frac{14 	imes 10^{-2}}{3.5} = 4 	imes 10^{-2} \ m = 0.04 \ m = 4 \ cm$.When the capillary is inclined at an angle $\phi = 60^{\circ}$ with the vertical,the length $l$ of the water column along the capillary is given by $l = \frac{h}{\cos \phi}$.$l = \frac{0.04}{\cos 60^{\circ}} = \frac{0.04}{0.5} = 0.08 \ m = 8 \ cm$.

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