• # question_answer An engine has an efficiency of $1/6$. When the temperature of sink is reduced by ${{62}^{o}}C,$ its efficiency is doubled. Temperatures of source and sink are A)  ${{99}^{o}}C,{{37}^{o}}C$          B)  ${{124}^{o}}C,{{62}^{o}}C$C)  ${{37}^{o}}C,{{99}^{o}}C$          D)  ${{62}^{o}}C,{{124}^{o}}C$

From $\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}},$$\frac{{{T}_{2}}}{{{T}_{1}}}=1-\eta =1-\frac{1}{6}=\frac{5}{6}$ ?(i) In 2nd case: $\frac{{{T}_{2}}-62}{{{T}_{1}}}=1-\eta '=1-\frac{2}{6}=\frac{2}{3}$   ?(ii) Using (i), ${{T}_{2}}-62=\frac{2}{3}{{T}_{1}}=\frac{2}{3}\times \frac{6}{5}{{T}_{2}}=\frac{4}{5}{{T}_{2}}$ or $\frac{1}{5}{{T}_{2}}=62,\,{{T}_{2}}=310K=310K=310-273={{37}^{o}}C$ ${{T}_{1}}=\frac{6}{5}{{T}_{2}}=\frac{6}{5}\times 310=372K=372-273={{99}^{o}}C$
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