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(D) The height of capillary rise or depression is given by the formula: $h = \frac{2T \cos 	heta}{r ho g}$.For water: $h_w = 10 \ cm$,$	heta_w = 0

Vedclass · Accepted Answer

$1:6.5$

Vedclass · Answer

(D) The height of capillary rise or depression is given by the formula: $h = \frac{2T \cos 	heta}{r ho g}$.For water: $h_w = 10 \ cm$,$	heta_w = 0^o$,$ho_w = 1 \ g/cc$,$T_w$ is surface tension of water.For mercury: $h_m = -3.42 \ cm$ (depression),$	heta_m = 135^o$,$ho_m = 13.6 \ g/cc$,$T_m$ is surface tension of mercury.Taking the ratio: $\frac{h_w}{h_m} = \frac{T_w \cos 	heta_w}{T_m \cos 	heta_m} 	imes \frac{ho_m}{ho_w}$.Substituting the values: $\frac{10}{-3.42} = \frac{T_w \cos 0^o}{T_m \cos 135^o} 	imes \frac{13.6}{1}$.Since $\cos 0^o = 1$ and $\cos 135^o = -\frac{1}{\sqrt{2}} \approx -0.707$,we have:$\frac{10}{-3.42} = \frac{T_w}{T_m 	imes (-0.707)} 	imes 13.6$.$\frac{T_w}{T_m} = \frac{10 	imes (-0.707)}{-3.42 	imes 13.6} = \frac{-7.07}{-46.512} \approx 0.152$.$0.152 \approx \frac{1}{6.58}$. Thus,the ratio is approximately $1:6.5$.

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