1. Power dissipation in the original wire

Given:

• Resistance of wire, $R = 36\,\Omega$
• Applied voltage, $V = 240\,\text{V}$

Power dissipated is $P = \frac{V^2}{R}$

So, $P_1 = \frac{240^2}{36}$

2. Resistance of each half after cutting

For a uniform wire, resistance is proportional to length. When the wire is cut into two equal halves, resistance of each half becomes $R' = \frac{R}{2} = \frac{36}{2} = 18\,\Omega$

3. Power dissipated by each half

Each half is connected separately across the same potential difference $240\,\text{V}$.

So power in each half is $P' = \frac{240^2}{18}$

4. Total power dissipated in the second case

There are two halves, each dissipating $\frac{240^2}{18}$. Thus, $P_2 = 2\times \frac{240^2}{18}$

5. Find the ratio

We need $P_1 : P_2 = \frac{240^2}{36} : 2\times \frac{240^2}{18}$

Cancel $240^2$: $= \frac{1}{36} : \frac{2}{18}$ $= \frac{1}{36} : \frac{1}{9}$

Now, $\frac{1/9}{1/36} = 4$

Hence, $P_1 : P_2 = 1 : 4$

So, $x = 4$.

6. Comparison with stored answer

Stored correct answer = $4$

This matches our derived answer.