JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2023
- Medium
- Numerical
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 \(.............\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(16\)
Step-by-step Solution
Detailed explanation
\(P=V I=I^2 R=\frac{V^2}{R}\)
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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