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

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The height of capillary rise is given by $h = \frac{2 T cos θ}{r ρ g}$

This implies $h \propto \frac{1}{r}$

The area of cross-section of the capillary tube is $a = π r^{2}$ , which gives $r \propto a$

Therefore, $h \propto \frac{1}{a}$

For the two capillary tubes, we can write $\frac{h _{2}}{h _{1}} = \frac{a _{1}}{a _{2}}$

Given $a_{1} = a$ and $a_{2} = 4 a$ , we get $\frac{h _{2}}{h} = \frac{a}{4 a} = \frac{1}{2}$

$h_{2} = \frac{h}{2}$

Answer: $\frac{h}{2}$

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