A) \[100{}^\circ C\]
B) \[76{}^\circ C\]
C) \[64{}^\circ C\]
D) \[50{}^\circ C\]
Correct Answer: C
Solution :
Given,\[\varepsilon =6+4t-\frac{{{t}^{2}}}{32}\] For neutral of thermocouple \[\frac{d\varepsilon }{dt}=0\] Here, \[\frac{d\varepsilon }{dt}=4-\frac{2t}{32}\] \[4-\frac{2t}{32}=0\] \[t=\frac{4\times 32}{2}\] \[t={{64}^{o}}C\]You need to login to perform this action.
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