JEE Physics 2023 Question with Solution

Given: The engine operates between the freezing point and boiling point of water.

Find: Which statements about efficiency are correct.

For a Carnot engine,

$$\eta = \left(1 - \frac{T_2}{T_1}\right) \times 100$$

where $T_1$ and $T_2$ are the temperatures of the hot and cold reservoirs in Kelvin.

Here,

$$T_1 = 100 + 273 = 373 \, \text{K}, \quad T_2 = 0 + 273 = 273 \, \text{K}$$

Substituting,

$$\eta = \left(1 - \frac{273}{373}\right) \times 100 = 26.8\%$$

So the Carnot efficiency between these temperatures is less than $27\%$.

Now, any real engine operating between the same two temperatures has efficiency less than the Carnot engine. Therefore statement (2) is true.

Since even the maximum possible efficiency here is $26.8\%$, statement (4) is true and statement (3) is false.

Thus the correct combination from the listed options should be $2 \text{ and } 4 \text{ only}$. However, the solution labels the correct option as B, while the options provided show $2 \text{ and } 3 \text{ only}$ as option B. This is inconsistent with the working. Based on the solution working, the defensible answer is the option containing $2 \text{ and } 4 \text{ only}$.

Given: A heat engine works between $373 \, \text{K}$ and $273 \, \text{K}$.

Find: Whether its efficiency can exceed, equal, or remain below $27\%$.

A Carnot engine gives the maximum possible efficiency for any engine operating between the same two reservoirs. Therefore,

$$\eta_{\text{real}} < \eta_{\text{Carnot}}$$

Now calculate the Carnot value:

$$\eta_{\text{Carnot}} = \left(1 - \frac{273}{373}\right) \times 100 \approx 26.8\%$$

Hence any actual engine must satisfy

$$\eta_{\text{real}} < 26.8\% < 27\%$$

So statement (2) is correct and statement (4) is correct. Statement (1) is impossible, and statement (3) is also false.

Therefore the correct option by working is C if the options are read as given here.