MCQ
Efficiency of engine is $\eta_1$ at $T_1 = 200^\circ C$ and $T_2 = 0^\circ C$ and $\eta_2$ at $T_1 = 0^\circ C$ and $T_2 = 200^\circ K.$ Find the ratio of $\frac{\eta_1}{\eta_2}.$
- A$1.00$
- B$0.721$
- ✓$0.577$
- D$0.34$
✓
Answer
Correct option: C.
$0.577$
$\eta_1=1-\frac{273+0}{200+273}=\frac{200}{473}$
$\eta_2=1-\frac{-200+273}{0+27 3}=\frac{200}{273}$
$\frac{\eta_1}{\eta_2\text{}}=\frac{200}{473}\times\frac{273 }{200}=\frac{273}{473}=0.577$
$\eta_2=1-\frac{-200+273}{0+27 3}=\frac{200}{273}$
$\frac{\eta_1}{\eta_2\text{}}=\frac{200}{473}\times\frac{273 }{200}=\frac{273}{473}=0.577$
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