Question

The radii of the two columns in a $U$ tube are $r_1$ & $r_2$, when a liquid of density (angle of contact is $0^o$)is filled in it, the level difference of the liquid in the two arms in $h$. The surface tension of the liquid is : ($g=$ acceleration due to gravity )

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

Answer 1

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

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

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

so, $h_1=\frac{2T}{r_1\rho g}-------\left(1\right)$

and $h_2=\frac{2T}{r_2\rho g}-------\left(2\right)$

Level differance of liquid in the two arms is given by

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

$\Rightarrow h=\frac{2T}{\rho g}[\frac{1}{r_1}-\frac{1}{r_2}]$

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

Hence,

$T=\frac{h\rho g{r}_1{r}_2}{2(r_2-r_1)}$