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(C) The height of a liquid column in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{rdg}$.Here,$T$ is surface tension,$	heta$ i

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$3$

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(C) The height of a liquid column in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{rdg}$.Here,$T$ is surface tension,$	heta$ is the angle of contact,$r$ is the radius of the capillary,$d$ is the density of the liquid,and $g$ is the acceleration due to gravity.Since the capillary tube is the same,$r$ and $g$ are constant. Thus,$h \propto \frac{T \cos 	heta}{d}$.For the first liquid (water): $h_1 = 6 	ext{ cm}$,$T_1 = 70 	ext{ dynes/cm}$,$	heta_1 = 0^\circ$,$d_1 = 1 	ext{ g/cm}^3$.For the second liquid: $T_2 = 140 	ext{ dynes/cm}$,$	heta_2 = 60^\circ$,$d_2 = 2 	ext{ g/cm}^3$.Using the ratio: $\frac{h_2}{h_1} = \frac{T_2}{T_1} 	imes \frac{\cos 	heta_2}{\cos 	heta_1} 	imes \frac{d_1}{d_2}$.Substituting the values: $\frac{h_2}{6} = \frac{140}{70} 	imes \frac{\cos 60^\circ}{\cos 0^\circ} 	imes \frac{1}{2}$.$\frac{h_2}{6} = 2 	imes \frac{0.5}{1} 	imes 0.5 = 2 	imes 0.25 = 0.5$.Therefore,$h_2 = 6 	imes 0.5 = 3 	ext{ cm}$.

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