A) 1127 K
B) 1400 K
C) 1154 K
D) 1100 K
Correct Answer: A
Solution :
Using relation \[{{R}_{t}}={{R}_{0}}\,(1+\alpha t)\] where \[{{R}_{0}}=\] resistance at \[{{0}^{o}}C\] \[{{R}_{t}}=\] resistance at \[{{t}^{o}}C\] \[\alpha =\] temperature coefficient = 0.00125 Here : \[{{t}_{1}}=300-273={{27}^{o}}C\] \[\therefore \] \[\frac{{{R}_{2}}}{{{R}_{1}}}=\frac{1+\alpha {{t}_{2}}}{1+\alpha {{t}_{1}}}\] \[2=\frac{1+0.00125\times {{t}_{2}}}{1+0.00125\times {{27}^{o}}}\] \[\Rightarrow \] \[0.00125\,{{t}_{2}}=2.0.675-1\] or \[{{t}_{2}}=\frac{1.0675}{0.00125}={{854}^{o}}C=1127K\]
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