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(C) The formula for capillary rise or fall is given by $h = \frac{2 \sigma \cos 	heta}{r ho g}$.From this,we can write $\sigma = \frac{h r ho g}{

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$1 : 6$

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(C) The formula for capillary rise or fall is given by $h = \frac{2 \sigma \cos 	heta}{r ho g}$.From this,we can write $\sigma = \frac{h r ho g}{2 \cos 	heta}$.Since the capillary tube is the same,$r$ is constant. Thus,$\sigma \propto \frac{h ho}{\cos 	heta}$.For water: $h_w = 10 \, cm$,$ho_w = 1 \, g/cm^3$,$	heta_w = 0^{\circ}$.For mercury: $h_m = -3.1 \, cm$ (fall),$ho_m = 13.6 \, g/cm^3$,$	heta_m = 135^{\circ}$.The ratio of surface tensions is $\frac{\sigma_w}{\sigma_m} = \frac{h_w ho_w}{\cos 	heta_w} 	imes \frac{\cos 	heta_m}{h_m ho_m}$.Substituting the values: $\frac{\sigma_w}{\sigma_m} = \frac{10 	imes 1}{\cos 0^{\circ}} 	imes \frac{\cos 135^{\circ}}{-3.1 	imes 13.6}$.Since $\cos 0^{\circ} = 1$ and $\cos 135^{\circ} = -\frac{1}{\sqrt{2}} \approx -0.707$,we get:$\frac{\sigma_w}{\sigma_m} = \frac{10 	imes (-0.707)}{-3.1 	imes 13.6} = \frac{-7.07}{-42.16} \approx \frac{1}{6}$.

Water rises to a height of $10 \, cm$ in a capillary tube and mercury falls to a depth of $3.1 \, cm$ in the same capillary tube. If the density of mercury is $13.6 \, g/cm^3$ and the angle of contact for mercury is $135^{\circ}$,the approximate ratio of surface tensions of water and mercury is

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