MHTCET 16 Apr 2026 Shift 2 Paper

Concept:Carnot engine efficiency depends only on source and sink temperatures, given by the relation $\eta = 1 - \frac{T_2}{T_1}$.Explanation:Let $T_1$ be the source temperature and $T_2$ be the sink temperature.Initial efficiency is $\eta = 50\% = 0.5$ and $T_1 = 600\ \mathrm{K}$.Using the Carnot formula:$$0.5 = 1 - \frac{T_2}{600}$$$$\frac{T_2}{600} = 0.5$$$$T_2 = 300\ \mathrm{K}$$So the sink temperature is $300\ \mathrm{K}$.The efficiency is increased by $20\%$, meaning new efficiency:$$\eta' = 0.5 + 0.2 = 0.7$$The sink temperature remains unchanged at $T_2 = 300\ \mathrm{K}$.Let the new source temperature be $T_1'$. Then:$$0.7 = 1 - \frac{300}{T_1'}$$$$\frac{300}{T_1'} = 0.3$$$$T_1' = \frac{300}{0.3} = 1000\ \mathrm{K}$$Answer:The new temperature of the source is $1000\ \mathrm{K}$.Correct option: C.