A potential $\mathrm{V}_{0}$ 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 $P_{2}: \mathrm{P}_{1}$ is $\sqrt{x}: 1$. The value of $x$ is ___________.

Let's analyze the initial situation where the potential $V_0$ is applied across the entire length of the wire with resistance $R$. <br/><br/> The power dissipation, $P_1$, can be calculated using the formula: <br/><br/> $P_1 = \frac{V_0^2}{R}$ <br/><br/> Now, let's consider the case where the wire is cut into two equal halves. Each half will have half the original resistance, $\frac{R}{2}$. The potential $V_0$ is applied across the length of each half. <br/><br/> For each half of the wire, the power dissipation, $P'$, can be calculated using the formula: <br/><br/> $P' = \frac{V_0^2}{\frac{R}{2}} = \frac{2V_0^2}{R}$ <br/><br/> Since there are two halves of the wire, the total power dissipation across the two wires, $P_2$, is: <br/><br/> $P_2 = 2P' = 2\left(\frac{2V_0^2}{R}\right) = \frac{4V_0^2}{R}$ <br/><br/> Now, let's find the ratio $P_2 : P_1$: <br/><br/> $\frac{P_2}{P_1} = \frac{\frac{4V_0^2}{R}}{\frac{V_0^2}{R}} = 4$ <br/><br/> Comparing this to the given ratio $\sqrt{x} : 1$, we have: <br/><br/> $\frac{P_2}{P_1} = \sqrt{x}$ <br/><br/> So, $\sqrt{x} = 4$. <br/><br/> Squaring both sides, we get: <br/><br/> $x = 16$ <br/><br/> The value of $x$ is 16.