Answer
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Hint:We have to know the formula of efficiency of any steam engine. The formula of efficiency of a steam engine is \[\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}\] . Here \[\eta \] is the efficiency of a steam engine. . In this question we might be confused by the fourth option but we have to remember that efficiency is not dependent on fuels.
Complete step by step answer:
We know that \[\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}\] ,
Here \[\eta \] is the efficiency of a steam engine, and \[{T_2}\& {T_1}\] is equal to the sink temperature and the source temperatures respectively.
Now, \[{\eta _A} = 1 - \dfrac{{{T_2}}}{{{T_1}}}\] , here \[{\eta _A}\]is equal to the efficiency of the steam engine A.
\[{T_1}\] \[ = \] \[700k\], \[{T_2}\]\[ = \] \[350k\]
Therefore,
\[{\eta _A} = 1 - \dfrac{{{T_2}}}{{{T_1}}} \\
\Rightarrow{\eta _A} = 1 - \dfrac{{350}}{{700}} \\
\Rightarrow{\eta _A} = 1 - 0.5 \\
\Rightarrow{\eta _A} = 0.5\]
Again for the engine B, \[{\eta _B} = 1 - \dfrac{{{T_2}}}{{{T_1}}}\]
\[{T_1}\] \[ = \]\[650k\], \[{T_2}\]\[ = \] \[300k\] ,
Therefore,
\[{\eta _B} = 1 - \dfrac{{{T_2}}}{{{T_1}}} \\
\Rightarrow {\eta _B} = 1 - \dfrac{{300}}{{650}}\\
\Rightarrow {\eta _B} = 1 - 0.46\\
\Rightarrow {\eta _B} = 0.53\]
From this calculation we can say the efficiency of B is greater than the efficiency of A.Therefore, if two steam engines A and B, have their sources respectively at \[700k\] and \[650k\] . And their sinks at \[350k\] and \[300k\] .Then the efficiency of the steam engine B will be greater than the efficiency of the steam engine A.
Hence, option B is correct.
Note:We have to keep in our mind that efficiencies cannot be affected by the fuel present in the engine. We can get confused by the last option given in the question. And the second important thing is we have to be careful that the formula is \[{\eta _{}} = 1 - \dfrac{{{T_2}}}{{{T_1}}}\] not \[{\eta _{}} = 1 - \dfrac{{{T_1}}}{{{T_2}}}\]. Many people make these small mistakes. And also we have to remember which one is the source temperature and which one is the sink. . And also we have to remember which one is the source temperature and which one is the sink. All temperatures should be calculated in kelvin units.And the efficiency has no unit because it is the ratio of the two temperatures.
Complete step by step answer:
We know that \[\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}\] ,
Here \[\eta \] is the efficiency of a steam engine, and \[{T_2}\& {T_1}\] is equal to the sink temperature and the source temperatures respectively.
Now, \[{\eta _A} = 1 - \dfrac{{{T_2}}}{{{T_1}}}\] , here \[{\eta _A}\]is equal to the efficiency of the steam engine A.
\[{T_1}\] \[ = \] \[700k\], \[{T_2}\]\[ = \] \[350k\]
Therefore,
\[{\eta _A} = 1 - \dfrac{{{T_2}}}{{{T_1}}} \\
\Rightarrow{\eta _A} = 1 - \dfrac{{350}}{{700}} \\
\Rightarrow{\eta _A} = 1 - 0.5 \\
\Rightarrow{\eta _A} = 0.5\]
Again for the engine B, \[{\eta _B} = 1 - \dfrac{{{T_2}}}{{{T_1}}}\]
\[{T_1}\] \[ = \]\[650k\], \[{T_2}\]\[ = \] \[300k\] ,
Therefore,
\[{\eta _B} = 1 - \dfrac{{{T_2}}}{{{T_1}}} \\
\Rightarrow {\eta _B} = 1 - \dfrac{{300}}{{650}}\\
\Rightarrow {\eta _B} = 1 - 0.46\\
\Rightarrow {\eta _B} = 0.53\]
From this calculation we can say the efficiency of B is greater than the efficiency of A.Therefore, if two steam engines A and B, have their sources respectively at \[700k\] and \[650k\] . And their sinks at \[350k\] and \[300k\] .Then the efficiency of the steam engine B will be greater than the efficiency of the steam engine A.
Hence, option B is correct.
Note:We have to keep in our mind that efficiencies cannot be affected by the fuel present in the engine. We can get confused by the last option given in the question. And the second important thing is we have to be careful that the formula is \[{\eta _{}} = 1 - \dfrac{{{T_2}}}{{{T_1}}}\] not \[{\eta _{}} = 1 - \dfrac{{{T_1}}}{{{T_2}}}\]. Many people make these small mistakes. And also we have to remember which one is the source temperature and which one is the sink. . And also we have to remember which one is the source temperature and which one is the sink. All temperatures should be calculated in kelvin units.And the efficiency has no unit because it is the ratio of the two temperatures.
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