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A Carnot engine with efficiency $\eta$ operates between two heat reservoirs with temperatures $T_1$ and $T_{2^{\prime}}$ where $T_1>T_2$. If only $T_1$ is changed by $0.4 \%$, the change in efficiency is $\Delta \eta_1$, whereas if only $T_2$ is changed by $0.2 \%$, the efficiency is changed by $\Delta \eta_2$. The ratio $\frac{\Delta \eta_1}{\Delta \eta_2}$ is approximately,
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Verified Answer
The correct answer is:
-2
Key Idea Efficiency of Carnot engine,
$$
\eta=1-\frac{T_2}{T_1}
$$
Where, $T_1=$ source temperature and $T_2=\sin k$ temperature
If $T_1$ is changed by $0.4 \%$, then
$$
\frac{\Delta \eta}{\eta}=\frac{\Delta T_1}{T_1}+\frac{\Delta T_2}{T_2}
$$
(From combination of the error)
$$
\Rightarrow \frac{\Delta \eta_1}{\eta}=\frac{0.4}{100}+0
$$
Similarly, $T_2$ is changed by $0.2 \%$, then
$$
\frac{\Delta \eta_2}{\eta}=0+\frac{0.2}{100}
$$
So, from Eq. (i) and (ii), we get
$$
\frac{\Delta \eta_1}{\Delta \eta_2}=\frac{0.4}{0.2}=2
$$
Hence, option (1) is correct.
$$
\eta=1-\frac{T_2}{T_1}
$$
Where, $T_1=$ source temperature and $T_2=\sin k$ temperature
If $T_1$ is changed by $0.4 \%$, then
$$
\frac{\Delta \eta}{\eta}=\frac{\Delta T_1}{T_1}+\frac{\Delta T_2}{T_2}
$$
(From combination of the error)
$$
\Rightarrow \frac{\Delta \eta_1}{\eta}=\frac{0.4}{100}+0
$$
Similarly, $T_2$ is changed by $0.2 \%$, then
$$
\frac{\Delta \eta_2}{\eta}=0+\frac{0.2}{100}
$$
So, from Eq. (i) and (ii), we get
$$
\frac{\Delta \eta_1}{\Delta \eta_2}=\frac{0.4}{0.2}=2
$$
Hence, option (1) is correct.
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