A) \[{{124}^{\text{o}}}C\]
B) \[{{37}^{\text{o}}}C\]
C) \[{{62}^{\text{o}}}C\]
D) \[{{99}^{\text{o}}}C\]
Correct Answer: D
Solution :
Efficiency of engine is given by \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] \[\therefore \frac{{{T}_{2}}}{{{T}_{1}}}=1-\eta =1-\frac{1}{6}=\frac{5}{6}....(i)\] In other case, \[\frac{{{T}_{2}}-62}{{{T}_{1}}}=1-\eta =1-\frac{2}{6}=\frac{2}{3}......(ii)\] Using Eq. (i), \[{{T}_{2}}-62=\frac{2}{3}\,{{T}_{1}}=\frac{2}{3}\times \frac{6}{5}{{T}_{2}}\] \[or\frac{1}{5}{{T}_{2}}=62\] \[\therefore {{T}_{2}}=310\,K\,=310-{{273}^{o}}C\] \[={{37}^{o}}C\] Hence, \[{{T}_{1}}=\frac{6}{5}{{T}_{2}}=\frac{6}{5}\times 310\] = 372K = 372-273 \[={{99}^{\text{o}}}C\] Hence, temperature of source is \[{{99}^{\text{o}}}C\].
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