A) \[\text{1}.\text{42 }\times \text{ 1}{{0}^{\text{3}}}\text{ V}\]
B) \[\text{1}.\text{42 }\times \text{ 1}{{0}^{-\text{3}}}\text{ V}\]
C) \[\text{14}.\text{2 }\times \text{ 1}{{0}^{\text{3}}}\text{ V}\]
D) \[\text{14}.\text{2 }\times \text{ 1}{{0}^{-\text{3}}}\text{ V}\]
Correct Answer: B
Solution :
\[{{E}_{Ni-Cu}}=aT+\frac{b}{2}{{T}^{2}}\] \[=(1.63\times {{10}^{-6}})\times (100)-(0.42\times {{10}^{-6}}){{(100)}^{2}}\] \[\therefore \] \[{{E}_{Ni-Cu}}=(1.63\times {{10}^{-4}})-2.1\times {{10}^{-4}}\] \[=14\times 2\times {{10}^{-4}}=1.42\times {{10}^{-3}}volt\].You need to login to perform this action.
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