Question

The radii of the two columns in 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 and $r_1<r_1$)

• 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 $\frac{\rho gh{r}_1{r}_2}{2(r_2-r_1)}$


We know,
$h=\frac{2T\cos \theta }{r\rho g}$
For limb $1$,
$h_1=\frac{2T\cos 0^{\circ }}{r_1\rho g}=\frac{2T}{r_1\rho g}$
Similarly,
For limb 2,
$h_2=\frac{2T\cos 0^{\circ }}{r_2\rho g}=\frac{2T}{r_2\rho g}$
Given,
$h_1-h_2=h$
$\therefore \frac{2T}{r_1\rho g}-\frac{2T}{r_2\rho g}=h$
$\Rightarrow \frac{2T}{\rho g}(\frac{1}{r_1}-\frac{1}{r_2})=h$
$\Rightarrow \frac{2T}{\rho g}\left(\frac{r_2-r_1}{r_1{r}_2}\right)=h$
$\Rightarrow T=\frac{h\rho g{r}_1{r}_2}{2(r_2-r_1)}$