JEE MAIN - Physics (2023 - 13th April Morning Shift - No. 23)
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 ___________.
Answer
16
Explanation
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.
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.
Comments (0)
