A) \[999.9{}^\circ C\]
B) \[1005.3{}^\circ C\]
C) \[1020.2{}^\circ C\]
D) \[1037.1{}^\circ C\]
Correct Answer: D
Solution :
(d) \[1037.1{}^\circ C\] Here, \[{{\text{R}}_{0}}=99\,\Omega ,{{T}_{0}}=27{}^\circ C\] \[{{\text{R}}_{T}}=116\,\Omega ;\,\alpha =1.7\times {{10}^{-4}}\,{}^\circ {{C}^{-1}}\] \[\therefore {{R}_{T}}={{R}_{0}}[1+\alpha (T-{{T}_{0}})]\] \[\therefore \frac{{{R}_{T}}}{{{R}_{0}}}-1=\alpha (T-{{T}_{0}})\] \[\Rightarrow \frac{116}{99}-1=\alpha \,(T-{{T}_{0}})\] \[T={{T}_{0}}=\frac{1}{\alpha }\left[ \frac{116-99}{99} \right]=\frac{17}{99\alpha }=\frac{1}{1.7\times {{10}^{-4}}}\times \frac{17}{99}\] \[\therefore T-{{T}_{0}}=\frac{{{10}^{5}}}{99}=1010.10{}^\circ C\] \[\Rightarrow T=1010.1+{{T}_{0}}=1010.1+27=1037.1{}^\circ C\]
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