JEE Main 2025 Jan 28 Shift 1, Physics Q7: Heat Engine and Carnot’s Cycle

JEE Main2025Jan 28, Shift 1Physics

A Carnot engine (E) is working between two temperatures 473K and 273K. In a new system two engines — engine $E_{1}$ works between 473K to 373K and engine $E_{2}$ works between 373K to 273K. If $η_{12}$ , $η_{1}$ and $η_{2}$ are the efficiencies of the engines $E$ , $E_{1}$ and $E_{2}$ , respectively, then

A
$η_{12} < η_{1} + η_{2}$
B
$η_{12} = η_{1} η_{2}$
C
$η_{12} = η_{1} + η_{2}$
D
$η_{12} \geq η_{1} + η_{2}$

Solution

Carnot efficiency: $η = 1 - \frac{T _{c}}{T _{h}}$ .

$η_{12} = 1 - \frac{273}{473} = \frac{200}{473} \approx 0.423$ .

$η_{1} = 1 - \frac{373}{473} = \frac{100}{473} \approx 0.211$ .

$η_{2} = 1 - \frac{273}{373} = \frac{100}{373} \approx 0.268$ .

$η_{1} + η_{2} \approx 0.479$ , which is greater than $η_{12} \approx 0.423$ .

$∴ η_{12} < η_{1} + η_{2}$ .

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JEE Main 2025 Jan 28 Shift 1, Physics Q7: Heat Engine and Carnot’s Cycle

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