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 $P_2: P_1$ is $\sqrt{x}: 1$. The value of $x$ is $.............$.
JEE MAIN 2023, Medium
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$P=V I=I^2 R=\frac{V^2}{R}$
Now $R=\frac{\rho l}{A}$
If wire is cut in two equal half
$R^{\prime}=\frac{R}{2}$
Initial $P_1=\frac{V_0^2}{R}$
After $P_2=\frac{V_0^2}{R^{\prime}} \times 2 \Rightarrow \frac{V_0^2}{R} \times 4$
$\frac{P_2}{P_1}=4=\frac{\sqrt{x}}{1}$
$x=16$
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