Skip to content
dronacharya.ai
← JEE Advanced 2026 · Paper 1
PhysicsNumerical+4Question 26 of 48

JEE Advanced 2026 · Paper 1, Physics question 10

As shown in the figure, five Carnot engines, each with efficiency η\eta and same number of cycles per unit time, are operating between six heat reservoirs. The amount of heat released per cycle by one engine is completely absorbed by the next engine. Consider Q0Q_0 to be the amount of heat absorbed per cycle by the first engine and WW as the amount of total work done by all the engines per cycle, then the net efficiency of the system is found to be ηnet=WQ0=211243\eta_{\text{net}}=\dfrac W{Q_0}=\dfrac{211}{243}. The value of η\eta is:

A vertical chain of six reservoirs (horizontal bars) with an engine (circle) between each adjacent pair. The first engine takes Q₀ from the top reservoir, gives work W₁ and rejects Q₁ to the next; the second takes Q₁, gives W₂ and rejects Q₂; dots indicate the chain continues to the fifth engine, which takes Q₄, gives W₅ and rejects Q₅ to the bottom reservoir.
Answer and worked solution

Official answer: 0.32

Each engine turns a fraction η\eta of its input into work and passes on 1−η1-\eta:
Qi=(1−η)iQ0,W=Q0−Q5=Q0[1−(1−η)5].Q_i=(1-\eta)^iQ_0,\qquad W=Q_0-Q_5=Q_0\left[1-(1-\eta)^5\right].

So
1−(1−η)5=211243  ⟹  (1−η)5=32243=(23)5,1-(1-\eta)^5=\frac{211}{243}\implies(1-\eta)^5=\frac{32}{243}=\left(\frac23\right)^5,
giving η=13≈0.33\eta=\frac13\approx0.33.

Topics:Carnot cycleHeat engines and refrigerators
← Previous questionNext question →