Hint: The power consumption depends on the resistivity of the wires.
Step 1:
Power consumption in a day i.e., in 5 hours = 10 units
Or power consumption per hour = 2 units
Or power consumption per hour = 2 units = 2 kW = 2000 J/s
Also, we know that power consumption in resistor,
P = V x l
⇒ 2000W = 220V x l or l = 9A
Step 2: Now, the resistance of the wire is given by R= ρlA where A is the cross-sectional area of the conductor.
Power consumption in the first current-carrying wire is given by;
P=l2R=ρlAI2=1.7×10−8×10π×(10−3)2×81 J/s≈ 4 J/s
Step 3: The fractional loss due to the joule heating in the first wire =42000×100=0.2%
Power loss in Al wire =P×ρAlρCu=1.6×4=6.4J/s
The fractional loss due to the joule heating in the second wire =6.42000×100=0.32%