A) \[0R\,\,\Omega \]
B) \[2R\,\,\Omega \]
C) \[3R\,\,\Omega \]
D) \[4R\,\,\Omega \]
Correct Answer: D
Solution :
\[{{R}_{t}}={{R}_{0}}(1+\alpha \,\,\Delta t)\] \[5R={{R}_{0}}(1+\alpha \times 50)\] ... (i) and \[6R={{R}_{0}}(1+\alpha \times 100)\] ... (ii) On dividing Eq. (i) by (ii) and solving it, we get \[\alpha =\frac{1}{200}\Omega /{{0}^{o}}C\] On putting in Eq. (i), we get \[5R={{R}_{0}}\left( 1+\frac{1}{4} \right)\] \[\Rightarrow \] \[5R=\frac{5}{4}{{R}_{0}}\] \[\Rightarrow \] \[{{R}_{0}}=4R\Omega \]
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