Explanation:
For a capillary tube,
\(rh = \) constant
where, \(r=\) radius of capillary tube
and \(h=\) height of rised water in capillary tube According to the question
\(r_{1} h_{1}=r_{2} h_{2}\)
\( \Rightarrow \frac{{{r_1}}}{{{r_2}}} = \frac{{{h_2}}}{{{h_1}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
In the first condition,
\({A_1} = \pi r_1^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\)
In the second condition,
\({A_2} = \pi r_2^2{\text{ }}\)
\({\rm{or }}\frac{A}{9} = \pi r_2^2\left[ {{\rm{ as}},{\rm{ }}{A_2} = \frac{A}{9}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\)
On dividing eq.(2) by (3), we get
\(9=\dfrac{r_{1}^{2}}{r_{2}^{2}} \operatorname{or} \dfrac{r_{1}}{r_{2}}=\sqrt{9} \Rightarrow \dfrac{r_{1}}{r_{2}}=3\)
from eq.(1), we get
\(\dfrac{h_{2}}{h_{1}}=3 \Rightarrow h_{2}=3 h_{1}\)
\(h_{2}=3 h\)
