Q138 / 160 · AP EAMCET 2016 · Physics

When the absolute temperature of the source of a Carnot heat engine is increased by $25 \%$, its efficiency increases by $80 \%$. The new efficiency of the engine is

  • A
    $12 \%$
  • B
    $24 \%$
  • C
    $48 \%$
  • Correct answer
    D
    $36 \%$

Solution

According to the question, initially $\begin{aligned} & \eta=\frac{100-T_2}{100} \\ & =\eta+\eta \times \frac{80}{100}=1.8 \eta=\frac{125-T_2}{125} \\ & \frac{\eta}{\eta^{\prime}}=\frac{100-T_2}{100} \times \frac{125}{125-T_2} \\ & \Rightarrow \quad \frac{1}{18}=\frac{100-T_2}{100} \times \frac{125}{125-T_2} \\ & \text { or, } \frac{5}{9}=\frac{100-T_2}{100} \times \frac{5}{\left(125-T_2\right)} \\ & \text { or, } 9\left(100-\mathrm{T}_2\right)=500-4 \mathrm{~T}_2 \\ & \text { or, } 900-9 \mathrm{~T}_2=500-4 \mathrm{~T}_2 \\ & \text { or, } 400=5 \mathrm{~T}_2 \\ & \text { or, } \mathrm{T}_2=\frac{400}{5}=80 \end{aligned}$ The new efficiency, $\begin{aligned} & \eta^{\prime \prime}=1-\frac{T_2}{T_1}=1-\frac{80}{125} \\ & \frac{125-80}{125}=\frac{45}{125}=\frac{45}{125} \times 100=36 \% \end{aligned}$

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

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