Liquid rises to a height 2 cm in a capillary | vedclass

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Question

(C) The height of the liquid column in a capillary tube is given by $h = \frac{2T \cos 	heta}{r ho g}$.When the tube is long enough,the liquid rise

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) The height of the liquid column in a capillary tube is given by $h = \frac{2T \cos 	heta}{r ho g}$.When the tube is long enough,the liquid rises to height $h = 2 \ cm$ with an angle of contact $	heta = 0^{\circ}$.When the tube is cut or lowered such that its length above the liquid surface is $h' = 1 \ cm$,the liquid will rise to the top of the tube and adjust its angle of contact to $	heta'$ to maintain equilibrium.Since $T$,$r$,$ho$,and $g$ are constant,we have $h \cos 	heta = h' \cos 	heta'$.Substituting the values: $2 	imes \cos(0^{\circ}) = 1 	imes \cos(	heta')$.Since $\cos(0^{\circ}) = 1$,we get $2 	imes 1 = \cos(	heta')$,which implies $\cos(	heta') = 0.5$.Therefore,$	heta' = \cos^{-1}(0.5) = 60^{\circ}$.

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