Q161 / 200 · MHT CET - Medical 2026 (25 April Shift 2) · Physics
Let $\eta_1$ be the efficiency of a Carnot engine at $T_H = 547^\circ\text{C}$ and $T_C = 247^\circ\text{C}$ while $\eta_2$ is the efficiency at $T_H = 847^\circ\text{C}$ and $T_C = 147^\circ\text{C}$. If $\eta_1 = \dfrac{x}{41}\eta_2$, the value of $x$ is
- Correct answerA$24$
- B$25$
- C$20$
- D$22$
Solution
The temperatures for the first Carnot engine in Kelvin are:
$T_{H1} = 547^{\circ}\text{C} + 273 = 820\text{ K}$
$T_{C1} = 247^{\circ}\text{C} + 273 = 520\text{ K}$
The efficiency $\eta_1$ is given by:
$\eta_1 = 1 - \dfrac{T_{C1}}{T_{H1}} = 1 - \dfrac{520}{820} = 1 - \dfrac{26}{41} = \dfrac{15}{41}$
The temperatures for the second Carnot engine in Kelvin are:
$T_{H2} = 847^{\circ}\text{C} + 273 = 1120\text{ K}$
$T_{C2} = 147^{\circ}\text{C} + 273 = 420\text{ K}$
The efficiency $\eta_2$ is given by:
$\eta_2 = 1 - \dfrac{T_{C2}}{T_{H2}} = 1 - \dfrac{420}{1120} = 1 - \dfrac{3}{8} = \dfrac{5}{8}$
Given the relation $\eta_1 = \dfrac{x}{41}\eta_2$, substituting the values of $\eta_1$ and $\eta_2$:
$\dfrac{15}{41} = \dfrac{x}{41} \times \dfrac{5}{8}$
$15 = \dfrac{5x}{8}$
$x = \dfrac{15 \times 8}{5} = 24$
Answer: $24$
$T_{H1} = 547^{\circ}\text{C} + 273 = 820\text{ K}$
$T_{C1} = 247^{\circ}\text{C} + 273 = 520\text{ K}$
The efficiency $\eta_1$ is given by:
$\eta_1 = 1 - \dfrac{T_{C1}}{T_{H1}} = 1 - \dfrac{520}{820} = 1 - \dfrac{26}{41} = \dfrac{15}{41}$
The temperatures for the second Carnot engine in Kelvin are:
$T_{H2} = 847^{\circ}\text{C} + 273 = 1120\text{ K}$
$T_{C2} = 147^{\circ}\text{C} + 273 = 420\text{ K}$
The efficiency $\eta_2$ is given by:
$\eta_2 = 1 - \dfrac{T_{C2}}{T_{H2}} = 1 - \dfrac{420}{1120} = 1 - \dfrac{3}{8} = \dfrac{5}{8}$
Given the relation $\eta_1 = \dfrac{x}{41}\eta_2$, substituting the values of $\eta_1$ and $\eta_2$:
$\dfrac{15}{41} = \dfrac{x}{41} \times \dfrac{5}{8}$
$15 = \dfrac{5x}{8}$
$x = \dfrac{15 \times 8}{5} = 24$
Answer: $24$
Hence, the correct answer is option: 'A'.