Question

Two capillary tube of radius $r_1$ and $r_2$ are put vertically one each in two liquids whose relative densities are $0.6$ and $0.4$ and surface tension are $60\text{dyne/cm}$ and $50\text{dyne/cm}$ respectively.
If ratio of heights of liquid in two tube is $8:10$ respectively. Find $\frac{r_1}{r_2}$.
(Assume contact angle is same in both case)

• A. $1$
• B. $\frac{1}{2}$
• C. $2$
• D. $\frac{2}{3}$

Answer 1

The correct option is A $1$

We know
$h=\frac{2T\cos \theta }{r\rho g}$
Where,
$h=$ Height of liquid in capillary
$r=$ Radius of capillary
For tube $1$
$h_1=\frac{2T_1\cos \theta }{r_1{\rho }_{1}g}$
$\Rightarrow r_1=\frac{2T_1\cos \theta }{h_1{\rho }_{1}g}$
Similarly,
For tube $2$
$r_2=\frac{2T_2\cos \theta }{h_2{\rho }_{2}g}$
$\therefore \frac{r_1}{r_2}=\frac{T_1}{T_2}\times \frac{h_2}{h_1}\times \frac{{\rho }_2}{{\rho }_1}=\frac{60}{50}\times \frac{10}{8}\times \frac{0.4}{0.6}$
$\Rightarrow \frac{r_1}{r_2}=1$