The temperature of the reservoir of a Carnot engine, operating with an efficiency of 70% is 1000K. The temperature of its sink is:

Correct option is A

​$$\textbf{Given:} \\\bullet \text{ Efficiency of Carnot engine, } \eta = 70\% = 0.7 \\\bullet \text{ Temperature of the source/reservoir, } T_1 = 1000\, \text{K}\\[10pt]\text{The efficiency of a Carnot engine is given by:} \\\eta = 1 - \frac{T_2}{T_1} \\\text{Where:} \\T_1 = \text{Temperature of the source (reservoir)} \\T_2 = \text{Temperature of the sink}\\[10pt]0.7 = 1 - \frac{T_2}{1000} \\\frac{T_2}{1000} = 1 - 0.7 = 0.3 \\T_2 = 0.3 \times 1000 = 300\, \text{K}$$​​