An electrical power line, having a total resistance of $$2\,\Omega$$, delivers $$1\,\text{kW}$$ at $$220\,\text{V}$$. The efficiency of the transmission line is approximately:

We are told that the line is supplying a useful power of $$P_{\text{out}} = 1\,\text{kW} = 1000\,\text{W}$$ to a load whose terminal voltage is $$V = 220\,\text{V}$$.

First, we find the current flowing through the line. We recall the electric‐power formula

$$P = VI,$$

where $$P$$ is the power delivered to the load, $$V$$ is the load voltage, and $$I$$ is the current. Solving this formula for $$I$$, we have

$$I = \frac{P}{V}.$$

Substituting $$P = 1000\,\text{W}$$ and $$V = 220\,\text{V}$$ gives

$$I = \frac{1000}{220}\,\text{A} \approx 4.545\,\text{A}.$$

Now, the transmission line itself has a total resistance of $$R = 2\,\Omega$$. Whenever current flows through a resistance, there is power lost as heat. The formula for this loss (Joule heating) is

$$P_{\text{loss}} = I^{2}R.$$

Substituting $$I \approx 4.545\,\text{A}$$ and $$R = 2\,\Omega$$, we obtain

$$\begin{aligned} P_{\text{loss}} &= (4.545\,\text{A})^{2}\times 2\,\Omega \\ &= 20.66 \times 2 \\ &\approx 41.32\,\text{W}. \end{aligned}$$

The total power that must be supplied at the sending end of the line is therefore the sum of the useful power delivered and the power lost:

$$P_{\text{in}} = P_{\text{out}} + P_{\text{loss}} = 1000\,\text{W} + 41.32\,\text{W} \approx 1041.32\,\text{W}.$$

Finally, the efficiency $$\eta$$ of the transmission line is defined as

$$\eta = \frac{P_{\text{out}}}{P_{\text{in}}}\times 100\%.$$

Substituting the numbers we have just found,

$$\eta = \frac{1000}{1041.32}\times 100\% \approx 0.9603 \times 100\% \approx 96\%.$$

So the efficiency of this transmission line is about $$96\%$$.

Hence, the correct answer is Option D.