A potential $$V_0$$ is applied across a uniform wire of resistance $$R$$. The power dissipation is $$P_1$$. The wire is then cut into two equal halves and a potential of $$V_0$$ is applied across the length of each half. The total power dissipation across two wires is $$P_2$$. The ratio of $$P_2 : P_1$$ is $$\sqrt{x} : 1$$. The value of $$x$$ is _____.

$$P = \frac{V^2}{R}$$ and $$R \propto L$$

$$P_1 = \frac{V_0^2}{R}$$

The wire is cut into two equal halves. Since $$R \propto L$$, the resistance of each half becomes $$R' = \frac{R}{2}$$.

$$P_{\text{half}} = \frac{V_0^2}{R/2} = 2 \left( \frac{V_0^2}{R} \right) = 2P_1$$

$$P_2 = P_{\text{half}} + P_{\text{half}} = 2P_1 + 2P_1 = 4P_1$$

$$P_2 : P_1$$ = $$4 : 1$$

$$x = 16$$