A) \[\frac{T}{5}\]
B) \[\frac{2T}{5}\]
C) \[6T\]
D) \[\frac{6T}{5}\]
Correct Answer: D
Solution :
The efficiency of a heat engine is defined as the ratio of work done to the heat supplied, i.e., \[\eta =\frac{Work\text{ }done}{Heat\text{ }input}=\frac{W}{Q}\] or \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] where \[{{T}_{2}}\] is temperature of sink, and \[{{T}_{1}}\]is temperature of hot reservoir. \[\therefore \] \[\frac{40}{100}=1-\frac{{{T}_{2}}}{{{T}_{1}}}\] \[\Rightarrow \] \[\frac{{{T}_{2}}}{{{T}_{1}}}0.6\] \[\Rightarrow \] \[{{T}_{1}}=0.6\,{{T}_{1}}\] Again, \[\frac{50}{100}=1-\frac{{{T}_{2}}}{{{T}_{1}}}\] \[\Rightarrow \] \[\frac{{{T}_{2}}}{{{T}_{1}}}=0.5\] \[\Rightarrow \] \[\frac{0.6\,{{T}_{1}}}{{{T}_{1}}}=0.5\] \[\therefore \] \[{{T}_{1}}=\frac{0.6}{0.5}{{T}_{1}}=\frac{6}{5}T\]
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