A) \[\frac{\rho gh{{r}_{1}}{{r}_{2}}}{2({{r}_{2}}-{{r}_{1}})}\]
B) \[\frac{\rho gh({{r}_{1}}-{{r}_{2}})}{2{{r}_{2}}{{r}_{1}}}\]
C) \[\frac{2({{r}_{1}}-{{r}_{2}})}{\rho gh{{r}_{1}}{{r}_{2}}}\]
D) \[\frac{\rho gh}{2({{r}_{2}}-{{r}_{1}})}\]
Correct Answer: A
Solution :
Let \[{{h}_{1}},{{h}_{2}}\] be the heights to which liquid rises in two columns of radii \[{{r}_{1}}\] and \[{{r}_{2}}\] respectively. Then \[{{h}_{1}}=\frac{25\cos 0{}^\circ }{{{r}_{1}}\rho g}=\frac{2S}{{{r}_{1}}\rho g}\] Where s is the surface tension of liquid. and \[{{h}_{2}}=\frac{2S\cos 0{}^\circ }{{{r}_{2}}\rho g}=\frac{2S}{{{r}_{2}}\rho g}\] \[\therefore \] Difference in levels of liquid in two arms of U tube is \[h={{h}_{1}}-{{h}_{2}}=\frac{2S}{\rho g}\left[ \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right]=2S\frac{({{r}_{2}}-{{r}_{1}})}{{{r}_{1}}{{r}_{2}}\rho g}\] \[S=\frac{{{r}_{1}}{{r}_{2}}\rho gh}{2({{r}_{2}}-{{r}_{1}})}\]
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