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) \[\frac{\rho gh}{2({{r}_{2}})-{{r}_{1}}}\]
Correct Answer: A
Solution :
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, \[T=\frac{h\rho g{{r}_{1}}{{r}_{2}}}{2({{r}_{2}}-{{r}_{1}})}\]You need to login to perform this action.
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