Question

The radii of the two columns is U-tube are $r_1$ and $r_2$ When a liquid of density $\rho$ (angle of contact is $0^{\circ }$) is filled in it the level difference of liquid in two arms is $h$ The surface tension of liquid is ($g$ acceleration due to gravity)

• A. $\frac{\rho gh{r}_1{r}_2}{2(r_2-r_1)}$
• B. $\frac{\rho gh(r_2-r_1)}{2r_1{r}_2}$
• C. $\frac{2(r_2-r_1)}{\rho gh{r}_1{r}_2}$
• D. $\frac{\rho gh}{2(r_2-r_1)}$

Answer 1

The correct option is A $\frac{\rho gh{r}_1{r}_2}{2(r_2-r_1)}$

As we know that

Rise of liquid in the capillary tube is given by $H=\frac{2T}{r\rho g}$

where, $T$ is the surface tension of the liquid, $r$ is the radius of the tube.

Given that $h_1$ is the height in the tube of radius $r_1$ and $h_2$ is the height in tube of radius of $r_2$

$⟹h_1=\frac{2T}{r_1\rho g}$

$⟹h_2=\frac{2T}{r_2\rho g}$

Level difference of liquid in the two arms is given by

$h=h_1-h_2=\frac{2T}{r_1\rho g}-\frac{2T}{r_2\rho g}$

$⟹h=\frac{2T}{\rho g}[\frac{r_2-r_1}{r_1{r}_2}]$

Hence Surface tension of the liquid, $T=\frac{h\rho g{r}_1{r}_2}{2(r_2-r_1)}$