A) 1154 K
B) 1100 K
C) 1400 K
D) 1127 K
Correct Answer: D
Solution :
(d) 1127 K Using, \[{{R}_{T}}={{R}_{0}}(1+\alpha T)\] \[\therefore \frac{{{R}_{{{T}_{2}}}}}{{{R}_{{{T}_{1}}}}}=\frac{{{R}_{0}}(1+\alpha {{T}_{2}})}{{{R}_{0}}(1+\alpha {{T}_{1}})}=\frac{2}{1}=\frac{(1+\alpha {{T}_{2}})}{(1+\alpha \times 300)}\] \[\Rightarrow 2+\alpha \times 600=1+\alpha {{T}_{2}}\] \[\Rightarrow 1=\alpha ({{T}_{2}}-600)\] \[\Rightarrow \frac{1}{0.00125}=({{T}_{2}}-600)\] \[\Rightarrow 800{}^\circ C={{T}_{2}}-600\] \[{{T}_{2}}=800-273+600\] \[{{T}_{2}}=1127\,K\]
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