Q75 / 90 · JEE Main 2022 (25 Jul Shift 2) · Physics

Let $η_{1}$ is the efficiency of an engine at $T_{1} = 447 ° C$ and $T_{2} = 147 ° C$ while $η_{2}$ is the efficiency at $T_{1} = 947 ° C$ and $T_{2} = 47 ° C$ . The ratio $\frac{η_{1}}{η_{2}}$ will be

A
$0.41$
Correct answer
B
$0.56$
C
$0.73$
D
$0.70$

Solution

The efficiency of an engine is given by $η = 1 - \frac{T_{L}}{T_{H}}$ . Here, $T_{L}$ and $T_{H}$ are in Kelvin scale.
$\Rightarrow η_{1} = 1 - \frac{147 + 273}{447 + 273} = 1 - \frac{420}{720}$
$\Rightarrow η_{1} = \frac{300}{720}$
And $η_{2} = 1 - \frac{47 + 273}{947 + 273} = 1 - \frac{320}{1220}$
$\Rightarrow η_{2} = \frac{900}{1220}$
$\Rightarrow \frac{η_{1}}{η_{2}} = \frac{300}{720} \times \frac{1220}{900} = \frac{122}{72 \times 3}$
$\Rightarrow \frac{η_{1}}{η_{2}} = 0.56$

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

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Let η1 is the efficiency of an engine at T1=447°C and T2=147°C while η2 is the efficiency at T1=947°C and T2=47°C. The ratio η1η2 will be

Solution

Relevant Questions

Let $η_{1}$ is the efficiency of an engine at $T_{1} = 447 ° C$ and $T_{2} = 147 ° C$ while $η_{2}$ is the efficiency at $T_{1} = 947 ° C$ and $T_{2} = 47 ° C$ . The ratio $\frac{η_{1}}{η_{2}}$ will be