A) \[36{}^\circ C\]
B) \[83{}^\circ C\]
C) \[63{}^\circ C\]
D) \[33{}^\circ C\]
E) \[66{}^\circ C\]
Correct Answer: B
Solution :
\[R={{R}_{0}}(1+\alpha t)\] \[\therefore \] \[{{R}_{0}}(1+30\alpha )=10\,\Omega \] and \[{{R}_{0}}(1+\alpha )=11\,\Omega \] Therefore, \[\frac{11}{10}=\frac{1+\alpha t}{1+30\alpha }\] Or \[11+330\alpha =10+10\alpha t\] Or \[11+330\times 0.002=10+10\times 0.002t\] Or \[11.66=0.02t+10\] Or \[0.02t=1.66\] Or \[t=83{}^\circ C\]
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