A) 1 ohm
B) 2 ohm
C) 3 ohm
D) 4 ohm
Correct Answer: D
Solution :
\[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{(1+\alpha {{t}_{1}})}{(1+\alpha {{t}_{2}})}\Rightarrow \frac{5}{6}=\frac{(1+\alpha \times 50)}{(1+\alpha \times 100)}\]\[\Rightarrow \alpha =\frac{1}{200}\text{ per}{{\text{ }}^{o}}C\] Again by \[{{R}_{t}}={{R}_{0}}(1+\alpha t)\] \[\Rightarrow \,\,5={{R}_{0}}\left( 1+\frac{1}{200}\times 50 \right)\Rightarrow {{R}_{0}}=4\Omega .\]
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