A) \[4.545\times {{10}^{-4}}{{/}^{o}}C\]
B) \[4.545\times {{10}^{-3}}{{/}^{o}}C\]
C) \[4.545\times {{10}^{-2}}{{/}^{o}}C\]
D) \[4.545\times {{10}^{-5}}{{/}^{o}}C\]
Correct Answer: A
Solution :
We have, \[{{R}_{t}}={{R}_{0}}(1+\alpha t)\] \[{{R}_{20}}={{R}_{0}}(1+20\alpha )\] ??(i) \[{{R}_{35}}={{R}_{0}}(1+35\alpha )\] ??(ii) From Eqs. (i) and (ii) \[\frac{{{R}_{20}}}{{{R}_{35}}}=\frac{1+20\alpha }{1+35\alpha }\] \[\frac{25}{25.17}=\frac{1+20\alpha }{1+35\alpha }\] \[25+875\alpha =25.17+503.4\alpha \] \[371.6\alpha =0.17\] \[\alpha =\frac{0.17}{317.6}\] \[=4.574\times {{10}^{4}}{{/}^{o}}C\]
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