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 $h = \frac{2 T cos θ}{ρ g r}$

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

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

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

Using the relation for two different tubes, we get $h_{1} a_{1} = h_{2} a_{2}$

Substituting the given values, $h a = h^{'} 4 a$

$h a = h^{'} (2 a)$

$h^{'} = \frac{h}{2}$

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

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