The efficiency of Carnot engine is 50% and temperature of sink is 500K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of sink will be

Efficiency $\eta$ of a carnot engine is given by $\eta =1-\frac{T_2}{T_1}$, where T₁ is the temperature of the source and T₂ is the temperature of the sink.
Here T₂ = 500 K
$\therefore 0.5=1-\frac{500}{T_1}\Rightarrow T_1=1000K$
Now, $\eta =0.6=1-\frac{T_2}{1000}$ 
(T₂ is the new sink temperature) $\Rightarrow T_2^{\text{'}}$= 400 K.