Question

If two capillary tubes of radii $r_1$ and $r_2$ in the ratio $1:2$ are dipped vertically in water, then the ratio of capillary rises in the respective tubes is

• A. $1:4$
• B. $4:1$
• C. $1:2$
• D. $2:1$

Answer 1

The correct option is D $2:1$

Rise of liquid in a capillary tube is given by,
$h=\frac{2T\cos \theta }{r\rho g}$
$\Rightarrow h\propto \frac{1}{r}...\left(i\right)$
[since for a given liquid and same material of capillary tube, density $\left(\rho \right)$, the surface tension $\left(T\right)$, angle of contact $\left(\theta \right)$ will remain constant.
i.e $\frac{2T\cos \theta }{\rho g}=\text{constant}$]
From Eq.$\left(i\right)$,
$\Rightarrow \frac{h_1}{h_2}=\frac{r_2}{r_1}$
$\because \frac{r_1}{r_2}=\frac{1}{2}$
$\Rightarrow \frac{h_1}{h_2}=\frac{2}{1}$