A) 1154 K
B) 1100 K
C) 1400 K
D) 1127K
Correct Answer: D
Solution :
| Using \[{{R}_{T}}={{R}_{0}}\left( 1+\alpha T \right)\] |
| \[\therefore \,\,\,\frac{{{R}_{{{T}_{2}}}}}{{{R}_{{{T}_{1}}}}}=\frac{{{R}_{0}}\left( 1+\alpha {{T}_{2}} \right)}{{{R}_{0}}\left( 1+\alpha {{T}_{1}} \right)}=\frac{2}{1}=\frac{\left( 1+\alpha {{T}_{2}} \right)}{\left( 1+\alpha \times 300 \right)}\] |
| \[\Rightarrow \,2+\alpha \times 600=1+\alpha {{T}_{2}}\] |
| \[\Rightarrow \,1=\alpha \left( {{T}_{2}}-600 \right)\Rightarrow \frac{1}{0.00125}=\left( {{T}_{2}}-600 \right)\] |
| \[\Rightarrow \,\,\,800{}^\circ C={{T}_{2}}-600\] |
| \[{{T}_{2}}=800-273+600\] |
| \[{{T}_{2}}=1127\,K\] |
You need to login to perform this action.
You will be redirected in
3 sec
