Q121 / 150 · MHT CET 2026 (17 April Shift 2) · Physics

A carnot engine having efficiency $\dfrac{1}{6}$, operates between the source temperature $T_H$ and the sink temperature $T_C$. Its efficiency increases to $\dfrac{1}{3}$, when $T_C$ is decreased by 64 K. The temperatures $T_H$ and $T_C$ are repectively

  • A
    $374$ K , $310$ K
  • Correct answer
    B
    $384$ K , $320$ K
  • C
    $384$ K , $340$ K
  • D
    $320$ K , $256$ K

Solution

The efficiency of a Carnot engine is given by $\eta = 1 - \dfrac{T_C}{T_H}$.

Initially, $\eta = \dfrac{1}{6}$, so:

$\dfrac{1}{6} = 1 - \dfrac{T_C}{T_H}$

$\dfrac{T_C}{T_H} = 1 - \dfrac{1}{6} = \dfrac{5}{6}$

$T_C = \dfrac{5}{6} T_H$

When $T_C$ is decreased by $64 \text{ K}$, the new sink temperature is $T_C - 64$. The new efficiency is $\dfrac{1}{3}$.

$\dfrac{1}{3} = 1 - \dfrac{T_C - 64}{T_H}$

$\dfrac{T_C - 64}{T_H} = 1 - \dfrac{1}{3} = \dfrac{2}{3}$

$\dfrac{T_C}{T_H} - \dfrac{64}{T_H} = \dfrac{2}{3}$

Substitute $\dfrac{T_C}{T_H} = \dfrac{5}{6}$ into the equation:

$\dfrac{5}{6} - \dfrac{64}{T_H} = \dfrac{2}{3}$

$\dfrac{64}{T_H} = \dfrac{5}{6} - \dfrac{2}{3}$

$\dfrac{64}{T_H} = \dfrac{5 - 4}{6} = \dfrac{1}{6}$

$T_H = 64 \times 6 = 384 \text{ K}$

Now, find $T_C$:

$T_C = \dfrac{5}{6} T_H = \dfrac{5}{6} \times 384 = 5 \times 64 = 320 \text{ K}$

Therefore, $T_H = 384 \text{ K}$ and $T_C = 320 \text{ K}$.

Hence, the correct answer is option: 'B'.

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