KCET · Physics · Rotational Motion
The efficiency of a Carnot engine which operates between the two temperatures \( T_{1}=500 \mathrm{~K} \)
and \( T_{2}=300 \mathrm{~K} \) is
- A \( 50 \% \)
- B 1) \( 25 \% \)
- C \( 75 \% \)
- D \( 40 \% \)
Answer & Solution
Correct Answer
(D) \( 40 \% \)
Step-by-step Solution
Detailed explanation
Efficiency, \( \eta=\left(1-\frac{T_{2}}{T_{1}}\right) \times 100 \% \)
Given \( T_{1}=500 ; T_{2}=300 \mathrm{~K} \)
Therefore,
\( \eta=\left(1-\frac{300}{500}\right) \times 100=\frac{200}{500} \times 100=40 \% \)
Given \( T_{1}=500 ; T_{2}=300 \mathrm{~K} \)
Therefore,
\( \eta=\left(1-\frac{300}{500}\right) \times 100=\frac{200}{500} \times 100=40 \% \)
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