1. Power used by family, p = 8kW = 8000 W
As only 20% of solar energy can be converted to useful electrical energy, power to be supplied by solar energy
= (8000w/20%)
= 40000 w.
Solar energy is incident at a rate of 200 w/m2 hence Area needed.
\(\frac {400w}{200Wm^{-2}}\) = 200 m2.
2. The area needed is comparable to roof area of large sized house.