A Carnot engine has efficiency of 50%. If the temperature of sink is reduced by 40°C, its efficiency increases by 30%. The temperature of the source will be:

We need to find the temperature of the source of a Carnot engine. The efficiency of a Carnot engine is given by $$\eta = 1 - \frac{T_2}{T_1}$$ and it is stated that initially $$\eta = 50\% = 0.5$$. Therefore $$0.5 = 1 - \frac{T_2}{T_1}$$ implies $$\frac{T_2}{T_1} = 0.5$$ and hence $$T_2 = 0.5 \, T_1 \quad \cdots (1)$$.

When the sink temperature is reduced by 40°C (equal to 40 K), the efficiency increases by 30%. If this increase were additive, the new efficiency would be $$0.5 + 0.3 = 0.8$$ so that $$0.8 = 1 - \frac{T_2 - 40}{T_1}$$ giving $$\frac{T_2 - 40}{T_1} = 0.2$$. Substituting from (1) yields $$\frac{0.5 \, T_1 - 40}{T_1} = 0.2$$, leading to $$0.5 \, T_1 - 40 = 0.2 \, T_1$$ and hence $$0.3 \, T_1 = 40$$ so that $$T_1 = \frac{40}{0.3} = 133.3 \text{ K}$$, which does not match any of the given options.

Interpreting the 30% increase as a relative increase instead gives a new efficiency of $$0.5 \times 1.3 = 0.65$$. In that case $$0.65 = 1 - \frac{T_2 - 40}{T_1}$$ implies $$\frac{T_2 - 40}{T_1} = 0.35$$. Using $$T_2 = 0.5 \, T_1$$ again, we have $$\frac{0.5 \, T_1 - 40}{T_1} = 0.35$$, so $$0.5 \, T_1 - 40 = 0.35 \, T_1$$, giving $$0.15 \, T_1 = 40$$ and hence $$T_1 = \frac{40}{0.15} = 266.67 \text{ K} \approx 266.7 \text{ K}$$.

Therefore, the correct answer is Option C: 266.7 K.