1 MW of power is to be delivered from a power station to a town 10 km…

The correct answer is Option (1) → 40 kW

The power lost in the line is the heat produced in its resistance, \( I^{2}R \). The current in the line is fixed by the transmission voltage (10000 V at the power station), not by the 220 V used in the town.

    \( \displaystyle I = \frac{P}{V} = \frac{1\times10^{6}\,\mathrm{W}}{1\times10^{4}\,\mathrm{V}} = 100\,\mathrm{A} \)

The two wires together make a line of length 2 × 10 km = 20 km.

    \( \displaystyle A = \pi r^{2} = \pi \times (0.5\times10^{-2}\,\mathrm{m})^{2} \approx 7.85\times10^{-5}\,\mathrm{m^{2}} \)

    \( \displaystyle R = \frac{\rho L}{A} = \frac{1.7\times10^{-8}\,\Omega\,\mathrm{m} \times 2\times10^{4}\,\mathrm{m}}{7.85\times10^{-5}\,\mathrm{m^{2}}} \approx 4.33\,\Omega \)

    \( \displaystyle P_{\mathrm{loss}} = I^{2}R = (100\,\mathrm{A})^{2} \times 4.33\,\Omega \)

    \( \displaystyle P_{\mathrm{loss}} \approx 4.3\times10^{4}\,\mathrm{W} \approx 43\,\mathrm{kW} \)

The nearest option is 40 kW.

Note: The exact working gives a power loss of about 43 kW, which is not printed among the options. The answer given by NTA, 40 kW, is the nearest value; the other options (80 kW, 100 kW and 120 kW) are far from the calculated loss.