1. Initial power dissipation

Let the resistance of the original wire be $R$.

When connected across a potential difference $V = 220\,\text{V}$, the power dissipated is

$P_1 = \frac{V^2}{R}$

2. Resistance after cutting the wire into two equal pieces

Resistance is proportional to length, so when the wire is cut into two equal parts, each piece has resistance

$\frac{R}{2}$

3. Equivalent resistance of two equal pieces in parallel

Each piece has resistance $\frac{R}{2}$, and they are connected in parallel:

$R_{\text{eq}} = \frac{\frac{R}{2}}{2} = \frac{R}{4}$

(Equivalently, for two equal resistors $r$ in parallel, $R_{\text{eq}} = \frac{r}{2}$.)

4. New power dissipation

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

$P_2 = \frac{V^2}{R_{\text{eq}}} = \frac{V^2}{R/4} = 4\frac{V^2}{R}$

So,

$P_2 = 4P_1$

Therefore,

$\frac{P_2}{P_1} = 4$

Hence,

$P_2 : P_1 = 4 : 1$

5. Option check

• A: $1$ ❌
• B: $4$ ✅
• C: $2$ ❌
• D: $3$ ❌

So the correct option is B.