QJEEM23Phfaba: JEE Main 2023 Physics - Class 12 JEEM23Phfaba Solution: A potential is applied across a uniform wire of resistance .…

Let's analyze the initial situation where the potential $V_0$ is applied across the entire length of the wire with resistance $R$ .

The power dissipation, $P_1$ , can be calculated using the formula:

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

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.

For each half of the wire, the power dissipation, $P'$ , can be calculated using the formula:

$P' = \frac{V_0^2}{\frac{R}{2}} = \frac{2V_0^2}{R}$

Since there are two halves of the wire, the total power dissipation across the two wires, $P_2$ , is:

$P_2 = 2P' = 2\left(\frac{2V_0^2}{R}\right) = \frac{4V_0^2}{R}$

Now, let's find the ratio $P_2 : P_1$ :

$\frac{P_2}{P_1} = \frac{\frac{4V_0^2}{R}}{\frac{V_0^2}{R}} = 4$

Comparing this to the given ratio $\sqrt{x} : 1$ , we have:

$\frac{P_2}{P_1} = \sqrt{x}$

So, $\sqrt{x} = 4$ .

Squaring both sides, we get:

$x = 16$

The value of $x$ is 16.