CBSE · Class 11 · Physics · ThermodynamicsThe efficiency ($\eta$) of a Carnot engine working between source temperature $T1$ and sink temperature $T2$ (in Kelvin) is given by:
Step-by-Step Solution
The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ is source temperature and $T_2$ is sink temperature in Kelvin.
Detailed Options Breakdown
Option : $1 - \frac{T_2}{T_1}$ (Correct Answer)
Correct choice. Refer to the step-by-step verified solution guidelines above for details.
Option 1: $1 - \frac{T_1}{T_2}$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Thermodynamics.
Option 2: $\frac{T_1}{T_2} - 1$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Thermodynamics.
Option 3: $1 + \frac{T_2}{T_1}$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Thermodynamics.
💡 Study Guide: This question tests core syllabus concepts from Thermodynamics. For formulas, key summaries, and mock exam reference guides, read the full Thermodynamics Revision Notes.