1. Initial power dissipation

When a potential difference $V$ is applied across a wire of resistance $R$, the power dissipated is

$W = \frac{V^2}{R}.$

2. Resistance of each half

If the wire is cut into two equal halves, resistance is proportional to length, so each half has resistance

$R_{\text{half}} = \frac{R}{2}.$

3. Equivalent resistance in parallel

Now these two halves, each of resistance $\frac{R}{2}$, are connected in parallel.

For two equal resistors $r$ in parallel, equivalent resistance is $\frac{r}{2}$.

So here,

$R_{\text{eq}} = \frac{1}{2}\left(\frac{R}{2}\right) = \frac{R}{4}.$

4. New power dissipation

Across the same supply voltage $V$, the new power is

$P' = \frac{V^2}{R_{\text{eq}}} = \frac{V^2}{R/4} = \frac{4V^2}{R}.$

But

$\frac{V^2}{R} = W,$

therefore

$P' = 4W.$

5. Option check

• A: $\frac{1}{2}W$ — incorrect
• B: $4W$ — correct
• C: $\frac{1}{4}W$ — incorrect
• D: $2W$ — incorrect

Hence, the energy dissipation rate becomes $4W$.