A uniform heating wire of resistance 36 $$\Omega$$ is connected across a potential difference of 240 V. The wire is then cut into half and a potential difference of 240 V is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be 1 : $$x$$, where $$x$$ is _________.

We begin by recalling the formula for the electric power developed across a resistor. For a resistor of resistance $$R$$ connected to a potential difference $$V$$, the power dissipated is given by the Joule-Lenz law:

$$P=\dfrac{V^{2}}{R}.$$

Now the original heating wire has a resistance of $$36\;\Omega$$ and is connected directly across a potential difference of $$240\;{\rm V}$$. Substituting these values in the power formula we get

$$P_{1}= \dfrac{(240)^{2}}{36}.$$

Next we cut the wire into two equal halves. Because the wire is uniform, its resistance is proportional to its length. Halving the length therefore halves the resistance. Thus the resistance of each half becomes

$$R_{\text{half}} = \dfrac{36\;\Omega}{2}=18\;\Omega.$$

Each half-wire is now connected independently across the same potential difference of $$240\;{\rm V}$$. Applying the power formula to one half, we obtain

$$P_{\text{half}} = \dfrac{(240)^{2}}{18}.$$

There are two such identical halves, so the total power dissipated in the second arrangement is the sum of the powers in the two halves:

$$P_{2} = 2\,P_{\text{half}} = 2\left(\dfrac{(240)^{2}}{18}\right)=\dfrac{(240)^{2}}{9}.$$

We are asked for the ratio of the power in the first case to the total power in the second case. Writing the ratio explicitly,

$$\dfrac{P_{1}}{P_{2}} = \dfrac{\dfrac{(240)^{2}}{36}}{\dfrac{(240)^{2}}{9}}.$$

The common factor $$(240)^{2}$$ cancels out, leaving

$$\dfrac{P_{1}}{P_{2}} = \dfrac{1/36}{1/9}= \dfrac{9}{36}= \dfrac{1}{4}.$$

Hence the ratio $$P_{1}:P_{2}$$ is $$1:4$$, so the value of $$x$$ is $$4$$.

Hence, the correct answer is Option 4.