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(B) The height of liquid rise in a capillary tube is given by the formula:$h = \frac{2T \cos 	heta}{r ho g}$Where $T$ is surface tension,$	heta$ i

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(B) The height of liquid rise in a capillary tube is given by the formula:$h = \frac{2T \cos 	heta}{r ho g}$Where $T$ is surface tension,$	heta$ is the angle of contact,$r$ is the radius of the tube,$ho$ is the density of the liquid,and $g$ is the acceleration due to gravity.When the level of the liquid inside and outside the capillary tube is the same,the height of the rise $h$ is $0$.Substituting $h = 0$ into the formula:$0 = \frac{2T \cos 	heta}{r ho g}$Since $T$,$r$,$ho$,and $g$ are non-zero constants,we must have:$\cos 	heta = 0$This implies that $	heta = 90^{\circ}$.

If a capillary tube is dipped into a liquid and the levels of the liquid inside and outside are the same,then the angle of contact is ...... $^o$.

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