Physics · Thermodynamics

JEE Main 2025 — 28 January, Morning Shift — Question 55

A Carnot engine (E) is working between two temperatures 473 K and 273 K . In a new system two engines - engine E_1 works between 473 K to 373 K and engine E_2 works between 373 K to 273 K . If η_12, η_1 and η_2 are the efficiencies of the engines E,E_1 and E_2, respectively, then

Option A:
${{\eta }_{12}}<{{\eta }_{1}}+{{\eta }_{2}}$
Correct
Option B:
${{\eta }_{12}}={{\eta }_{1}}{{\eta }_{2}}$
Option C:
${{\eta }_{12}}={{\eta }_{1}}+{{\eta }_{2}}$
Option D:
${{\eta }_{12}}\ge {{\eta }_{1}}+{{\eta }_{2}}$

Answer: A

Step-by-step solution

$\text{ }\!\!~\!\!\text{ }{{\eta }_{12}}=1-\frac{273}{473}=\frac{200}{473}=0.423$ ${{\eta }_{1}}=1-\frac{373}{473}=\frac{100}{473}=0.211$ ${{\eta }_{2}}=1-\frac{273}{373}=\frac{100}{373}=0.268$

Answer key and solution verified before publishing.

Practise Thermodynamics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 28 January, Morning Shift
Subject: Physics
Chapter: Thermodynamics
Topic: Entropy, Carnot's Engines and Refrigerators