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(D) The height of a liquid column in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{rdg}$.Given that the diameter $(2r)$,contact

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$\frac{9}{10}$

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(D) The height of a liquid column in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{rdg}$.Given that the diameter $(2r)$,contact angle $(	heta)$,and acceleration due to gravity $(g)$ are constant for both tubes,the height $h$ is proportional to $\frac{T}{d}$,where $T$ is surface tension and $d$ is density.Therefore,the ratio of the heights is given by: $\frac{h_1}{h_2} = \frac{T_1}{T_2} 	imes \frac{d_2}{d_1}$.Substituting the given values: $T_1 = 60 \, dyne/cm$,$T_2 = 50 \, dyne/cm$,$d_1 = 0.8$,and $d_2 = 0.6$.$\frac{h_1}{h_2} = \frac{60}{50} 	imes \frac{0.6}{0.8} = \frac{6}{5} 	imes \frac{6}{8} = \frac{6}{5} 	imes \frac{3}{4} = \frac{18}{20} = \frac{9}{10}$.

Two capillary tubes of same diameter are put vertically one each in two liquids whose relative densities are $0.8$ and $0.6$ and surface tensions are $60$ and $50 \, dyne/cm$ respectively. The ratio of the heights of the liquids in the two tubes $\frac{h_1}{h_2}$ is:

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