Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Heat and Thermodynamics question

2025 · 28 Jan · Shift 1 · Q70
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Heat and Thermodynamics
  5. /2025 · 28 Jan · Shift 1 · Q70

Heat and Thermodynamics question

2025 · 28 Jan · Shift 1 · Q70

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A Carnot engine (E)(\mathrm{E})(E) is working between two temperatures 473 K and 273 K . In a new system two engines - engine E1E_1E1​ works between 473 K to 373 K and engine E2E_2E2​ works between 373 K to 273 K . If η12,η1\eta_{12}, \eta_1η12​,η1​ and η2\eta_2η2​ are the efficiencies of the engines E,E1E, E_1E,E1​ and E2E_2E2​, respectively, then
  1. A
    η12=η1η2\eta_{12}=\eta_1 \eta_2η12​=η1​η2​
  2. B
    η12=η1+η2\eta_{12}=\eta_1+\eta_2η12​=η1​+η2​
  3. C
    η12≥η1+η2\eta_{12} \geq \eta_1+\eta_2η12​≥η1​+η2​
  4. D
    η12<η1+η2\eta_{12}\lt \eta_1+\eta_2η12​<η1​+η2​
View written solutionFree

Correct answer: D

  1. Carnot efficiency formula

For a Carnot engine working between temperatures THT_HTH​ and TCT_CTC​,

η=1−TCTH\eta = 1 - \frac{T_C}{T_H}η=1−TH​TC​​
  1. Efficiency of engine EEE

Engine EEE works between 473 K473\,\text{K}473K and 273 K273\,\text{K}273K.

η12=1−273473=473−273473=200473\eta_{12} = 1 - \frac{273}{473} = \frac{473-273}{473} = \frac{200}{473}η12​=1−473273​=473473−273​=473200​
  1. Efficiency of engine E1E_1E1​

Engine E1E_1E1​ works between 473 K473\,\text{K}473K and 373 K373\,\text{K}373K.

η1=1−373473=100473\eta_1 = 1 - \frac{373}{473} = \frac{100}{473}η1​=1−473373​=473100​
  1. Efficiency of engine E2E_2E2​

Engine E2E_2E2​ works between 373 K373\,\text{K}373K and 273 K273\,\text{K}273K.

η2=1−273373=100373\eta_2 = 1 - \frac{273}{373} = \frac{100}{373}η2​=1−373273​=373100​
  1. Check each option

Option A: η12=η1η2\eta_{12} = \eta_1\eta_2η12​=η1​η2​

η1η2=100473⋅100373=10000176429\eta_1\eta_2 = \frac{100}{473}\cdot \frac{100}{373} = \frac{10000}{176429}η1​η2​=473100​⋅373100​=17642910000​

This is clearly not equal to

η12=200473\eta_{12} = \frac{200}{473}η12​=473200​

So, A is false.

Option B: η12=η1+η2\eta_{12} = \eta_1 + \eta_2η12​=η1​+η2​

η1+η2=100473+100373\eta_1 + \eta_2 = \frac{100}{473} + \frac{100}{373}η1​+η2​=473100​+373100​

This is not equal to 200473\dfrac{200}{473}473200​.

So, B is false.

Compare η12\eta_{12}η12​ with η1+η2\eta_1+\eta_2η1​+η2​

Let us use the cascade relation for two engines in series:

If two engines operate sequentially, the combined efficiency is

η12=1−(1−η1)(1−η2)\eta_{12} = 1 - (1-\eta_1)(1-\eta_2)η12​=1−(1−η1​)(1−η2​)

Expanding,

η12=η1+η2−η1η2\eta_{12} = \eta_1 + \eta_2 - \eta_1\eta_2η12​=η1​+η2​−η1​η2​

Since η1η2>0\eta_1\eta_2 > 0η1​η2​>0,

η12<η1+η2\eta_{12} < \eta_1 + \eta_2η12​<η1​+η2​

Therefore, D is true and C is false.

  1. Numerical verification
η12=200473≈0.4228\eta_{12} = \frac{200}{473} \approx 0.4228η12​=473200​≈0.4228 η1=100473≈0.2114\eta_1 = \frac{100}{473} \approx 0.2114η1​=473100​≈0.2114 η2=100373≈0.2681\eta_2 = \frac{100}{373} \approx 0.2681η2​=373100​≈0.2681 η1+η2≈0.4795\eta_1 + \eta_2 \approx 0.4795η1​+η2​≈0.4795

Thus,

0.4228<0.47950.4228 < 0.47950.4228<0.4795

again confirming

η12<η1+η2\eta_{12} < \eta_1 + \eta_2η12​<η1​+η2​
  1. Final answer

The correct option is:

D\boxed{\text{D}}D​
PreviousNext

More from Heat and Thermodynamics

  • The ratio of vapour densities of two gases at the same temperature is 254​, then the ratio of r.m.s. velocities will be :2025 · MCQ
  • The kinetic energy of translation of the molecules in 50 g of CO2​ gas at 17°C is :2025 · MCQ
  • The work done in an adiabatic change in an ideal gas depends upon only :2025 · MCQ
  • A container of fixed volume contains a gas at 27°C. To double the pressure of the gas, the temperature of gas should be raised to ​ °C.2025 · Numerical
  • The difference of temperature in a material can convert heat energy into electrical energy. To harvest the heat energy, the material should have2025 · MCQ
  • Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in…2025 · MCQ
  • A cup of coffee cools from 90°C to 80°C in t minutes when the room temperature is 20°C. The time taken by the similar cup of coffee to cool from 80°C to 60°C at the same room temperature is:2025 · MCQ
  • A poly-atomic molecule (CV​=3R,CP​=4R, where R is gas constant) goes from phase space point A(PA​=105 Pa, VA​=4×10−6 m3) to… Includes diagram2025 · MCQ