Resistance is the opposition offered by a material to the flow of electric current. Resistivity, on the other hand, is the intrinsic property of a material that quantifies how strongly it resists the flow of electric current. The relationship between resistance (R), resistivity (ρ), and dimensions (length l, and cross-sectional area A) of the conductor is given by R=ρ⋅lA
Resistance is dependent on temperature, increasing with an increase in temperature for most conductors due to increased collision frequency of charge carriers with lattice ions, which impedes their flow.

The resistance of the bulb can be calculated using Ohm's law: R=V/I=330V/110W=3Ω The energy consumed by three bulbs burning for 5 hours is E=P⋅t=110W×3×5h=1650Wh=1.65kWh The cost in rupees is calculated as Cost=Energy consumed×Rate=1.65kWh×0.70 Rupees/kWh=1.155

The resistance of the copper wire can be calculated using the formula R=ρ⋅l/A, where ρρ is the resistivity of copper, ll is the length of the wire, and AA is the cross-sectional area of the wire. Given ρ=1.72×10−8 Ω⋅mρ=1.72×10−8Ω⋅m, l=2 km=2000 ml=2km=2000m, and r=2 mm=0.002 mr=2mm=0.002m (radius),the cross-sectional area A=πr2=π×(0.002)2 m2A=πr2=π×(0.002)2m2. Substituting these values into the formula gives R=(1.72×10−8 Ω⋅m)×(2000 m)π×(0.002)2 m2R=π×(0.002)2m2(1.72×10−8Ω⋅m)×(2000m). Calculating this expression gives the resistance of the wire.

• Domestic appliances typically use parallel connections. This is because in parallel connections, each appliance gets the full voltage of the power supply, ensuring consistent operation regardless of the other appliances in the circuit. Additionally, if one appliance fails or is turned off, it does not affect the operation of other appliances connected in parallel.

To calculate the energy consumed by the electric bulbs in the month of February, first, we calculate the energy consumed by the 250-watt bulb and the four 6-watt bulbs separately. Then, we sum up the total energy consumed. Energy consumed by the 250-watt bulb in 5 hours = 250×5/1000 kWh Energy consumed by each 6-watt bulb in 4.5 hours = 6×4.5×4/1000 kWh Total energy consumed in the month of February = Energy consumed by the 250-watt bulb + Energy consumed by the four 6-watt bulbs