1. The heat developed per unit time is the electrical power:

$P = \frac{V^2}{R}$

since the applied voltage $V$ is constant.

2. For a metallic wire,

$R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}$

where $L$ is length and $r$ is radius.

3. Initially, let the resistance be

$R_1 = \rho \frac{L}{\pi r^2}$

4. After the change:

• length is halved: $L' = \frac{L}{2}$
• radius is doubled: $r' = 2r$

So new area is

$A' = \pi (2r)^2 = 4\pi r^2$

Thus new resistance is

$R_2 = \rho \frac{L/2}{4\pi r^2} = \rho \frac{L}{8\pi r^2} = \frac{R_1}{8}$

5. Since voltage is constant,

$P_1 = \frac{V^2}{R_1}, \qquad P_2 = \frac{V^2}{R_2} = \frac{V^2}{R_1/8} = 8\frac{V^2}{R_1} = 8P_1$

6. Therefore, the rate of heat developed becomes 8 times the original value.

So the correct option is:

$\boxed{\text{D: Increased 8 times}}$

7. Comparison with stored answer:

• Stored correct answer: D
• Derived answer: D

They agree.