Two capillary tubes of radii 0.2 cm and 0.4 | vedclass

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(B) The height $h$ to which a liquid rises in a capillary tube is given by the formula $h = \frac{2T \cos 	heta}{r ho g}$,where $T$ is surface tens

Vedclass · Accepted Answer

$2:1$

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(B) The height $h$ to which a liquid rises 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.Since the same liquid is used in both tubes,$T$,$	heta$,$ho$,and $g$ are constant.Therefore,$h \propto \frac{1}{r}$,which implies $h_1 r_1 = h_2 r_2$.The ratio of the heights is $\frac{h_1}{h_2} = \frac{r_2}{r_1}$.Given $r_1 = 0.2 \ cm$ and $r_2 = 0.4 \ cm$,we have $\frac{h_1}{h_2} = \frac{0.4}{0.2} = \frac{2}{1}$.Thus,the ratio is $2:1$.

Two capillary tubes of radii $0.2 \ cm$ and $0.4 \ cm$ are dipped in the same liquid. The ratio of the heights to which the liquid will rise in the tubes is:

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