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(D) The efficiency of a reversible engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink tempera

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$99^\circ C, 37^\circ C$

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(D) The efficiency of a reversible engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature in Kelvin.Initially,$\eta = \frac{1}{6}$,so $1 - \frac{T_2}{T_1} = \frac{1}{6} \implies \frac{T_2}{T_1} = \frac{5}{6} \implies T_2 = \frac{5}{6}T_1$ ...$(i)$When the sink temperature is reduced by $62^\circ C$ (which is equivalent to $62 \ K$),the new efficiency $\eta' = 2\eta = 2 	imes \frac{1}{6} = \frac{1}{3}$.The new sink temperature is $T_2' = T_2 - 62$.Thus,$\eta' = 1 - \frac{T_2 - 62}{T_1} = \frac{1}{3}$.Substituting $T_2 = \frac{5}{6}T_1$ into the equation: $1 - \frac{\frac{5}{6}T_1 - 62}{T_1} = \frac{1}{3}$.$1 - (\frac{5}{6} - \frac{62}{T_1}) = \frac{1}{3} \implies 1 - \frac{5}{6} + \frac{62}{T_1} = \frac{1}{3}$.$\frac{1}{6} + \frac{62}{T_1} = \frac{1}{3} \implies \frac{62}{T_1} = \frac{1}{3} - \frac{1}{6} = \frac{1}{6}$.$T_1 = 62 	imes 6 = 372 \ K = (372 - 273)^\circ C = 99^\circ C$.Now,$T_2 = \frac{5}{6} 	imes 372 = 310 \ K = (310 - 273)^\circ C = 37^\circ C$.

$A$ reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by $62^\circ C$,the efficiency of the engine is doubled. The temperatures of the source and sink are:

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