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(D) Let the source temperature be $T_1$ and the sink temperature be $T_2$ (in Kelvin).Efficiency $\eta = 1 - \frac{T_2}{T_1} = \frac{1}{6}$.So,$\frac{

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

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(D) Let the source temperature be $T_1$ and the sink temperature be $T_2$ (in Kelvin).Efficiency $\eta = 1 - \frac{T_2}{T_1} = \frac{1}{6}$.So,$\frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} \implies T_2 = \frac{5}{6}T_1$.When the sink temperature is reduced by $62^{\circ} C$,the new sink temperature is $T_2' = T_2 - 62$.The new efficiency is doubled,so $\eta' = 2 	imes \frac{1}{6} = \frac{1}{3}$.Thus,$1 - \frac{T_2 - 62}{T_1} = \frac{1}{3} \implies \frac{T_2 - 62}{T_1} = \frac{2}{3}$.Substituting $T_2 = \frac{5}{6}T_1$: $\frac{\frac{5}{6}T_1 - 62}{T_1} = \frac{2}{3} \implies \frac{5}{6} - \frac{62}{T_1} = \frac{4}{6}$.$\frac{62}{T_1} = \frac{1}{6} \implies T_1 = 372 \ K$.Converting to Celsius: $T_1 = 372 - 273 = 99^{\circ} C$.$T_2 = \frac{5}{6} 	imes 372 = 310 \ K = 310 - 273 = 37^{\circ} C$.The temperatures are $99^{\circ} C$ and $37^{\circ} C$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X)$	$k^2$	$2k$	$k$	$2k$	$5k^2$

$A$ 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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