In capillary tube having area of cross-section ' $A$ ', water rises to a height ' $h$ '. If cross sectional…
In capillary tube having area of cross-section ' $A$ ', water rises to a height ' $h$ '. If cross sectional area is reduced to $\frac{A}{9}$, the rise of water in the capillary tube is
h
4 h
3 h
2h
Solution
$\begin{aligned} & h=\frac{2 T \cos \theta}{r \rho g} \\ & \Rightarrow h \propto \frac{1}{r} \\ & \text { And } A=\pi r^2 \Rightarrow r \propto \sqrt{A} \\ & \Rightarrow h \propto \frac{1}{\sqrt{A}} \\ & \text { Now, } \frac{h_2}{h_1}=\sqrt{\frac{A_1}{A_2}} \\ & \Rightarrow \frac{h_2}{h}=\sqrt{\frac{A}{A / 9}}=3 \\ & \Rightarrow h_2=3 h\end{aligned}$