A) \[16{}^\circ C\]
B) \[12{}^\circ C\]
C) \[8{}^\circ C\]
D) \[20{}^\circ C\]
Correct Answer: A
Solution :
Thermo-emf of thermocouple\[=25\mu V/{}^\circ C\] 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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