400oC
300oC
273oC
127oC
Correct answer is D
\(\frac{P_1}{T_1}\) = \(\frac{ P_2}{T_2}\)
\(\frac{3 \times\ 10^6}{27 + 273}\) = \(\frac{4 \times\ 10^6}{0 + 273}\)
\(\theta\) = 127oC
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