A) \[{{40}^{o}}C\]
B) \[{{50}^{o}}C\]
C) \[{{60}^{o}}C\]
D) \[{{70}^{o}}C\]
Correct Answer: C
Solution :
We have, \[{{R}_{t}}={{R}_{0}}(1+\alpha t)\] Here\[,\] \[3.70=2.71[1+\alpha (100-10)]\] ? (i) and \[3.26=2.71[1+\alpha (x-10)]\] ... (ii) From Eqs. (i) and (ii) \[\alpha =\frac{3.70-2.71}{90\times 2.71}\approx \frac{1}{90}\times 2.71\] ? (iii) Substituting Eq. (iii) in Eq. (i) \[3.26=2.71\left[ 1+\frac{1}{90\times 2.71}\times (x-10) \right]\] \[=2.71+\frac{x-10}{90}\] or \[\frac{x-10}{90}=3.26-2.71=0.55\] \[\Rightarrow \] \[x=(90\times 0.55)+10\] \[=49.5+10={{59.5}^{o}}C\] \[\approx {{60}^{o}}C\]
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