The Carnot efficiency is $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ is the source temperature and $T_2$ is the sink temperature (both in Kelvin). This is the maximum efficiency possible for any heat engine.
Back to JEE Main 2024 Thermodynamics
JEE Main 2024 — Physics Thermodynamics
hard
mcq
2024
Official previous-year question
Verified 30 May 2026.
Question
The efficiency of a Carnot engine operating between temperatures $T_1$ (source) and $T_2$ (sink) where $T_1 > T_2$ is:
Options
- A
$\eta = 1 - \frac{T_2}{T_1}$
- B
$\eta = 1 - \frac{T_1}{T_2}$
- C
$\eta = \frac{T_2}{T_1}$
- D
$\eta = T_1 - T_2$
Solution
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