Correct Answer: Option A

Explanation

85\(\Omega\) ........100^oC

52\(\Omega\) ........\(\theta\)^oC

20\(\Omega\) ........0^oC

\(\frac{52 - 20}{85 - 20} = \frac{\theta - 0}{100 - 0}\)

\(\theta = \frac{32 \times 100}{65} = 49.2^oC\)

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