A) \[{{16}^{o}}C\]
B) \[{{12}^{o}}C\]
C) \[{{8}^{o}}C\]
D) \[{{20}^{o}}C\]
Correct Answer: A
Solution :
| [a] Thermo-emf of thermocouple\[=25\mu \,V{{/}^{o}}C\]. |
| Let \[\theta \]be the smallest temperature difference. |
| Therefore, after connecting the thermocouple with the galvanometer, thermo-emf |
| \[E=(25\mu V{{/}^{o}}C)\times \theta {{(}^{o}}C)=25\theta \times {{10}^{-6}}V\] |
| Potential drop developed across the galvanometer |
| \[=iR={{10}^{-5}}\times 40=4\times {{10}^{-4}}V\] |
| \[\therefore \] \[4\times {{10}^{-4}}=25\theta \times {{10}^{-6}}\] |
| \[\therefore \] \[\theta =\frac{4}{25}\times {{10}^{2}}\] |
| \[={{16}^{o}}C\] |
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