A bulb with a rating of 250 V , 100 W is connected to a 220 V power...

The resistance of the bulb is given by:

$R = \dfrac{V^2}{P} = \dfrac{250^2}{100} = 625 \Omega$

Since the power supply is situated $10 \text{ m}$ away, a complete circuit requires both a forward and a return path. Thus, the total length of the connecting wire is:

$l = 2 \times 10 = 20 \text{ m}$

The resistance of the connecting wire is:

$R_w = \rho \dfrac{l}{A} = \dfrac{1.7 \times 10^{-8} \times 20}{5 \times 10^{-6}} = 0.068 \Omega$

The total resistance of the circuit is:

$R_{\text{eq}} = R + R_w = 625 + 0.068 \approx 625 \Omega$

The current flowing through the circuit is:

$I = \dfrac{V'}{R_{\text{eq}}} = \dfrac{220}{625} = 0.352 \text{ A}$

The power consumed by the connecting wires is:

$P_w = I^2 R_w = (0.352)^2 \times 0.068 \approx 0.1239 \times 0.068 \approx 0.0084 \text{ W}$