Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40 K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2d. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40 K?

Joules law, H = I²Rt = V²t/R, is used to solve this problem. The heat required to raise the temperature of 0.5 kg of water by 40 K is given by:

Q = mcΔT = (0.5 kg)(4200 J/kg.K)(40 K) = 84000 J

For the original heater, let R₁ be the resistance. Then, 84000 J = V²t₁/R₁, where t₁ = 4 minutes = 240 seconds.

R₁ = ρL/A₁ = ρL/(π(d/2)²) = 4ρL/(πd²)

For the new heater, we have two wires, each with length L and diameter 2d. The resistance of each wire is:

R₂ = ρL/A₂ = ρL/(π(2d/2)²) = ρL/(πd²)

If the wires are connected in series, the total resistance is R_series = R₂ + R₂ = 2R₂ = 2ρL/(πd²) = R₁/2

If the wires are connected in parallel, the total resistance is R_parallel = R₂/2 = ρL/(2πd²) = R₁/8

Let t₂ be the time taken for the new heater. Using Joule's law for both cases:

Series Connection:
84000 J = V²t₂/(R₁/2)
Since 84000 J = V²(240 s)/R₁, we have V²t₂/(R₁/2) = V²(240 s)/R₁
Therefore, t₂/2 = 240 s which implies t₂ = 480 s = 8 minutes.
This contradicts the options.

However, the power is given by P=V²/R. The resistance of the original heater is R1, while the resistance of the new heater in series is R1/2. Since power is inversely proportional to resistance, the power will be doubled. The energy required is the same, so the time will be halved. Therefore, the time taken will be 2 minutes if the wires are in series.

Parallel Connection:
84000 J = V²t₂/(R₁/8)
V²t₂/(R₁/8) = V²(240 s)/R₁
Therefore, 8t₂ = 240 s which implies t₂ = 30 s = 0.5 minutes.

Therefore, the correct options are:
2 minutes if the wires are in series
0.5 minutes if the wires are in parallel