1. Heat required is same in all cases

To raise the temperature of the same mass of water by the same amount, $Q = mc\Delta T$ is unchanged.

Hence, time taken is inversely proportional to the electrical power of the heater: $t \propto \frac{1}{P}$

Also, for a fixed supply voltage, $P = \frac{V^2}{R}$ so $t \propto R$

Thus, if we find the new equivalent resistance, we can compare times.

2. Resistance of original heater

For a wire, $R = \rho \frac{L}{A}$ where $A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}$

So original resistance is $R_0 = \rho \frac{L}{\pi d^2/4} = \frac{4\rho L}{\pi d^2}$

Given time with this heater: $t_0 = 4 \text{ min}$

3. Resistance of each new wire

Each new wire has:

• length $L$
• diameter $2d$

So its cross-sectional area becomes $A' = \pi \left(\frac{2d}{2}\right)^2 = \pi d^2$

Thus resistance of each new wire is $R' = \rho \frac{L}{\pi d^2}$

Compare with $R_0$: $R_0 = \frac{4\rho L}{\pi d^2}$ so $R' = \frac{R_0}{4}$

Each new wire has resistance $\dfrac{R_0}{4}$.

4. Case 1: wires in series

Equivalent resistance: $R_s = R' + R' = 2\left(\frac{R_0}{4}\right) = \frac{R_0}{2}$

Since time is proportional to resistance, $t_s = t_0 \cdot \frac{R_s}{R_0} = 4 \cdot \frac{1}{2} = 2 \text{ min}$

So option B is correct.

5. Case 2: wires in parallel

Equivalent resistance: $R_p = \frac{R'}{2} = \frac{1}{2}\cdot \frac{R_0}{4} = \frac{R_0}{8}$

Hence, $t_p = t_0 \cdot \frac{R_p}{R_0} = 4 \cdot \frac{1}{8} = 0.5 \text{ min}$

So option D is correct.

6. Check all options

• A: 4 min in parallel $\to$ incorrect, actual is $0.5$ min.
• B: 2 min in series $\to$ correct.
• C: 1 min in series $\to$ incorrect, actual is $2$ min.
• D: 0.5 min in parallel $\to$ correct.

7. Final answer

Correct options are: $\boxed{B,\ D}$