A) \[124{}^\circ C\]
B) \[37{}^\circ C\]
C) \[62{}^\circ C\]
D) \[99{}^\circ C\]
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, \[{{T}_{1}}=\frac{6}{5}{{T}_{2}}=\frac{6}{5}\times 310\] |
| \[=\text{ }372\,K\] |
| \[=372-273\] |
| \[=99{}^\circ C\] |
| Hence, temperature of source is \[99{}^\circ C\]. |
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