In a capillary tube of area of cross-section | 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 surfac

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$\frac{h}{2}$

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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$' is the radius of the capillary tube,'$ho$' is the density of the liquid, and '$g$' is the acceleration due to gravity.From this formula, we see that $h \propto \frac{1}{r}$.The area of cross-section '$a$' is given by $a = \pi r^2$, which implies $r = \sqrt{\frac{a}{\pi}}$, so $r \propto \sqrt{a}$.Substituting this into the height relation, we get $h \propto \frac{1}{\sqrt{a}}$.Let $h_1 = h$ for area $a_1 = a$, and $h_2$ be the height for area $a_2 = 4a$.Then, $\frac{h_2}{h_1} = \sqrt{\frac{a_1}{a_2}} = \sqrt{\frac{a}{4a}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore, $h_2 = \frac{h}{2}$.

In a capillary tube of area of cross-section '$a$',water rises to a height '$h$'. To what height will water rise in a capillary tube of area of cross-section $4a$?

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