A) \[v_{f}^{2}=v_{i}^{2}+\frac{2G{{m}_{e}}}{{{R}_{e}}}\left( 1+\frac{1}{10} \right)\]
B) \[v_{f}^{2}=v_{i}^{2}+\frac{2G{{m}_{e}}}{{{R}_{e}}}\left( 1-\frac{1}{10} \right)\]
C) \[v_{f}^{2}=v_{i}^{2}+\frac{2Gm}{{{R}_{e}}}\left( 1-\frac{1}{10} \right)\]
D) \[\mathbf{\vec{v}}\]
Correct Answer: A
Solution :
It is found that temperature of inversion \[{{i}_{2}}=i=\frac{E}{{{R}_{1}}}\]is as much above the neutral temperature \[{{i}_{3}}=i=\frac{E}{{{R}_{1}}+{{R}_{2}}}\]as neutral temperature is above the temperature of the cold junction (T), ie, \[{{i}_{3}}=i=\frac{E}{{{R}_{1}}+{{R}_{2}}}\] or \[{{T}_{n}}={{T}_{i}}+T\] But, here the cold junction is kept at \[\beta \] hence \[\Omega \]. Thus, \[\Omega \] or \[\Omega \]
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