Question

The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $0{}^\circ$ ) is filled in it, the level difference of the liquid in the two arms is 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. $\frac{\rho gh({{r}_{1}}-{{r}_{2}})}{{{r}_{2}}{{r}_{1}}}$
• C. $\frac{2({{r}_{1}}-{{r}_{2}})}{\rho gh{{r}_{1}}{{r}_{2}}}$
• D. $\frac{\rho gh}{2({{r}_{2}})-{{r}_{1}}}$

Answer 1

The Correct Option is A

Rise of liquid in the capillary tube is given by $h=\frac{2T}{r\rho g}$ Suppose, ${{h}_{1}}$ is the height in tube of radius ${{r}_{1}}$ and ${{h}_{2}}$ is the height in tube of radius of ${{r}_{2}}$ So, ${{h}_{1}}=\frac{2T}{{{r}_{1}}\rho g}$ ?(i) and ${{h}_{2}}=\frac{2T}{{{r}_{2}}\rho g}$ ?(ii) Level difference of liquid in the two arms is given by $h={{h}_{1}}-{{h}_{2}}=\frac{2T}{{{r}_{1}}\rho g}-\frac{2T}{{{r}_{2}}\rho g}$ $\Rightarrow$ $h=\frac{2T}{\rho g}\left[ \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right]$ $\Rightarrow$ $h=\frac{2T}{\rho g}\left[ \frac{{{r}_{2}}-{{r}_{1}}}{{{r}_{1}}{{r}_{2}}} \right]$ Hence, $T=\frac{h\rho g{{r}_{1}}{{r}_{2}}}{2({{r}_{2}}-{{r}_{1}})}$