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}})} $