Correct Answer: Option B

Explanation

100^oC .... 1.05\(\theta\)

\(\theta\)^oC .... 0.9\(\theta\)

0^oC .... 0.75\(\theta\)

\(\frac{\theta - 0}{100 - 0}\) = \(\frac{0.9 - 0.75}{1.05 - 0.75}\)

\(\theta\) = 50^oC

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