A Carnot engine (E) is working between two temperatures 473 K and 273 K . In a new system two engines - engine $$E_{1}$$ works between 473 K to 373 K and engine $$E_{2}$$ works between 373 K to 273 K . If $$\eta_{12},\eta_{1}$$ and $$\eta_{2}$$ are the efficiencies of the engines $$E,E_{1}$$ and $$E_{2}$$, respectively, then

The efficiency of a Carnot engine is given by the formula:

$$\eta = 1 - \frac{T_c}{T_h}$$

where $$T_c$$ is the cold reservoir temperature and $$T_h$$ is the hot reservoir temperature, both in Kelvin.

For the original engine $$E$$ working between $$T_1 = 473 \, \text{K}$$ and $$T_2 = 273 \, \text{K}$$:

$$\eta_{12} = 1 - \frac{T_2}{T_1} = 1 - \frac{273}{473} = \frac{473 - 273}{473} = \frac{200}{473}$$

For engine $$E_1$$ working between $$T_1 = 473 \, \text{K}$$ and $$T_3 = 373 \, \text{K}$$:

$$\eta_1 = 1 - \frac{T_3}{T_1} = 1 - \frac{373}{473} = \frac{473 - 373}{473} = \frac{100}{473}$$

For engine $$E_2$$ working between $$T_3 = 373 \, \text{K}$$ and $$T_2 = 273 \, \text{K}$$:

$$\eta_2 = 1 - \frac{T_2}{T_3} = 1 - \frac{273}{373} = \frac{373 - 273}{373} = \frac{100}{373}$$

Now, evaluate the options:

Option A: $$\eta_{12} = \eta_1 \eta_2$$

Compute $$\eta_1 \eta_2$$:

$$\eta_1 \eta_2 = \left( \frac{100}{473} \right) \times \left( \frac{100}{373} \right) = \frac{10000}{473 \times 373}$$

Compare with $$\eta_{12} = \frac{200}{473}$$:

$$\frac{200}{473} \neq \frac{10000}{473 \times 373}$$

Since they are not equal, option A is incorrect.

Option B: $$\eta_{12} \geq \eta_1 + \eta_2$$

Compute $$\eta_1 + \eta_2$$:

$$\eta_1 + \eta_2 = \frac{100}{473} + \frac{100}{373} = 100 \left( \frac{1}{473} + \frac{1}{373} \right) = 100 \left( \frac{373 + 473}{473 \times 373} \right) = 100 \times \frac{846}{473 \times 373} = \frac{84600}{473 \times 373}$$

Now, express $$\eta_{12}$$ with the same denominator:

$$\eta_{12} = \frac{200}{473} = \frac{200 \times 373}{473 \times 373} = \frac{74600}{473 \times 373}$$

Compare $$\frac{74600}{473 \times 373}$$ and $$\frac{84600}{473 \times 373}$$:

Since $$74600 < 84600$$, it follows that $$\eta_{12} < \eta_1 + \eta_2$$.

Thus, $$\eta_{12} \geq \eta_1 + \eta_2$$ is false, so option B is incorrect.

Option C: $$\eta_{12} = \eta_1 + \eta_2$$

From above:

$$\eta_{12} = \frac{200}{473} \approx 0.4228$$

$$\eta_1 + \eta_2 = \frac{100}{473} + \frac{100}{373} \approx 0.2114 + 0.2681 = 0.4795$$

Since $$0.4228 \neq 0.4795$$, they are not equal. Thus, option C is incorrect.

Option D: $$\eta_{12} \leq \eta_1 + \eta_2$$

From the comparison in option B, $$\eta_{12} < \eta_1 + \eta_2$$, which implies $$\eta_{12} \leq \eta_1 + \eta_2$$ is true.

Therefore, option D is correct.

The correct answer is option D.