MHTCET 17 Apr 2026 Shift 1 Paper

Concept:The efficiency of a Carnot engine is given by $$\eta = 1 - \frac{T_2}{T_1}$$ where $T_1$ is the source (intake) temperature and $T_2$ is the sink (exhaust) temperature.Explanation:Step 1: Find the exhaust temperature $T_2$ using the initial values.Given, initial efficiency $\eta_1 = 40\% = 0.40$ and source temperature $T_1 = 600\ \mathrm{K}$.So, $$0.40 = 1 - \frac{T_2}{600}$$ $$\frac{T_2}{600} = 1 - 0.40 = 0.60$$ $$T_2 = 0.60 \times 600 = 360\ \mathrm{K}$$ Step 2: For the same exhaust temperature $T_2 = 360\ \mathrm{K}$, the new efficiency is $\eta_2 = 60\% = 0.60$.Let the new intake temperature be $T_1'$.$$0.60 = 1 - \frac{360}{T_1'}$$ $$\frac{360}{T_1'} = 1 - 0.60 = 0.40$$ $$T_1' = \frac{360}{0.40} = 900\ \mathrm{K}$$Answer:The required intake temperature is $900\ \mathrm{K}$.Hence, the correct option is B. 900 K.