Let $\eta_{1}$ is the efficiency of an engine at $T _{1}=447^{\circ}\,C$ and $T _{2}=147^{\circ}\,C$ while $\eta_{2}$ is the efficiency at $T _{1}=947^{\circ}\,C$ and $T _{2}=47^{\circ}\,C$. The ratio $\frac{\eta_{1}}{\eta_{2}}$ will be.
JEE MAIN 2022, Medium
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Efficiency $\eta=1-\frac{ T _{ L }}{ T _{ H }}$
$\eta_{1}=1-\frac{147+273}{447+273}=1-\frac{420}{720}$
$\eta_{1}=\frac{300}{720}$
$\eta_{2}=1-\frac{47+273}{947+273}=1-\frac{320}{1220}$
$\eta_{2}=\frac{900}{1220}$
$\frac{\eta_{1}}{\eta_{2}}=\frac{300}{720} \times \frac{1220}{900}=\frac{122}{72 \times 3}$
$\frac{\eta_{1}}{\eta_{2}}=0.56$
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