A) \[\tan \phi =\frac{1}{2\tan \alpha }\]
B) \[\tan \lambda =2\tan \phi \]
C) \[\tan \lambda =2\tan \phi \]
D) \[\tan \phi =2\tan \lambda \]
Correct Answer: D
Solution :
For a dipole at position (R, Q) We have \[{{\text{p}}_{\text{1}}}\text{=10mm, }{{\text{p}}_{\text{2}}}\text{=50mm}\] ...(1) and \[{{C}_{6}}{{H}_{6}}\] ...(2) Also \[=\frac{10}{60}=\frac{1}{6}\] ...(3) Dividing Eq. (i) by Eq. (ii) \[=8\times \frac{1}{8}+6\times \frac{1}{2}=4\] ....(iv) From Eq. (iii) and Eq. (iv) \[=12\times \frac{1}{4}=3\] From figure, \[\text{C}{{\text{u}}_{\text{4}}}\text{A}{{\text{g}}_{\text{3}}}\text{Au}\text{.}\] So \[-nF{{E}^{\circ }}_{cell}=-RT\] \[{{E}^{\circ }}_{cell}=\frac{RT}{nF}\] \[{{E}^{\circ }}_{cell}\]You need to login to perform this action.
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