A) \[80{}^\circ \,C\]
B) \[70{}^\circ \,C\]
C) \[60{}^\circ \,C\]
D) \[50{}^\circ \,C\]
Correct Answer: A
Solution :
\[{{\operatorname{R}}_{20}}=\,\,20\,\Omega ,\,\,{{R}_{500}}=\,\,60\] \[{{\operatorname{R}}_{t}}=\,\,{{R}_{0}}^{(1\,\,+\,\,\alpha \,t)}\] \[\frac{{{R}_{20}}}{{{R}_{500}}}=\frac{{{R}_{0}}(1+\alpha \,\,\times 20)}{{{R}_{0}}(1+\alpha \,\,\times 500)}\] \[\Rightarrow \,\,\,\frac{20}{60}=\frac{1+20\alpha }{1+500\alpha }\,\,\Rightarrow \,\,1\,\,+\,\,500\,\alpha =3+60\,\alpha \] \[\Rightarrow \,\,\,\,500\,\alpha -60\,\alpha =\,\,2\,\,\Rightarrow \,\,\alpha =\frac{2}{440}\,\,=\,\,\frac{1}{220}\] \[\frac{{{R}_{20}}}{{{R}_{t}}}=\frac{{{R}_{0}}\left( 1+\frac{1}{220}\times 20 \right)}{{{R}_{0}}\left( 1+\frac{1}{220}\times t \right)}\] \[\frac{20}{25}=\frac{1+\frac{1}{11}}{1+\frac{t}{220}}\,\,\Rightarrow \,\,\frac{4}{5}=\frac{12/11}{1+t/220}\,\,\Rightarrow \,\,1+\frac{t}{220}=\frac{15}{11}\] \[\Rightarrow \,\,\,\operatorname{t}=\,\,80{}^\circ \,C.\]
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