A) \[40{{\,}^{o}}C\]
B) \[50{{\,}^{o}}C\]
C) \[60{{\,}^{o}}C\]
D) \[7\,0{{\,}^{o}}C\]
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) \[\alpha =\frac{3.70-2.71}{90\times 2.71}=\frac{1}{90}\times 2.71\] ?(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 \[\frac{x-10}{90}=3.26-2.71=0.55\] \[\Rightarrow \] \[x=(90\times 0.55)+10\] \[=49.5+10\] \[=59.5{{\,}^{o}}C\] \[=60{{\,}^{o}}C\]
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