A) 720 K
B) 345 K
C) 993 K
D) 690 K
Correct Answer: C
Solution :
Given\[{{R}_{300}}=0.3\Omega ,\,\,{{R}_{t}}=0.6\Omega \] \[T=300\,\,K={{27}^{o}}C\] Temperature coefficient of resistance, \[\alpha =1.5\times {{10}^{-3}}{{K}^{-1}}\] \[\therefore \] \[{{R}_{300}}={{R}_{0}}(1+\alpha \times 27)\] \[0.3={{R}_{0}}(1+1.5\times {{10}^{-3}}\times 27)\] ? (i) Again, \[{{R}_{t}}={{R}_{0}}(1+\alpha t)\] \[0.6={{R}_{0}}(1+1.5\times {{10}^{-3}}\times t)\] ... (ii) Dividing Eq. (ii) by Eq. (i), we get \[\frac{0.6}{0.3}=\frac{1+1.5\times {{10}^{-3}}t}{1+1.5\times {{10}^{-3}}\times 27}\] \[\Rightarrow \] \[2(1+1.5\times {{10}^{-3}}\times 27)=1+1.5\times {{10}^{-3}}t\] \[\Rightarrow \] \[2+81\times {{10}^{-3}}=1+1.5\times {{10}^{-3}}t\] \[\Rightarrow \] \[2+0.081=1+1.5\times {{10}^{-3}}t\] \[\Rightarrow \] \[t=\frac{1.081}{1.5\times {{10}^{-3}}}={{720}^{o}}C=993K\]
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