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Current Electricity question

2023 · 13 Apr · Shift 1 · Q67
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Current Electricity question

2023 · 13 Apr · Shift 1 · Q67

JEE MainPhysicsCurrent ElectricityNumerical+4 / −1
A potential V0\mathrm{V}_{0}V0​ is applied across a uniform wire of resistance RRR. The power dissipation is P1P_{1}P1​. The wire is then cut into two equal halves and a potential of V0V_{0}V0​ is applied across the length of each half. The total power dissipation across two wires is P2P_{2}P2​. The ratio P2:P1P_{2}: \mathrm{P}_{1}P2​:P1​ is x:1\sqrt{x}: 1x​:1. The value of xxx is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 16

  1. Initial wire

A wire of resistance RRR is connected across a potential difference V0V_0V0​.

So, the power dissipated is

P1=V02R.P_1 = \frac{V_0^2}{R}.P1​=RV02​​.
  1. Wire cut into two equal halves

For a uniform wire, resistance is proportional to length.

If the original resistance is RRR, then each half has resistance

R′=R2.R' = \frac{R}{2}.R′=2R​.
  1. Power dissipated in each half

A potential difference V0V_0V0​ is applied across each half-wire.

Hence power in each half is

Peach=V02R/2=2V02R.P_{\text{each}} = \frac{V_0^2}{R/2} = \frac{2V_0^2}{R}.Peach​=R/2V02​​=R2V02​​.
  1. Total power in two halves

Since there are two such halves,

P2=2×2V02R=4V02R.P_2 = 2 \times \frac{2V_0^2}{R} = \frac{4V_0^2}{R}.P2​=2×R2V02​​=R4V02​​.
  1. Ratio P2:P1P_2 : P_1P2​:P1​
P2P1=4V02/RV02/R=4.\frac{P_2}{P_1} = \frac{4V_0^2/R}{V_0^2/R} = 4.P1​P2​​=V02​/R4V02​/R​=4.

So,

P2:P1=4:1.P_2 : P_1 = 4:1.P2​:P1​=4:1.

Given,

P2:P1=x:1.P_2 : P_1 = \sqrt{x} : 1.P2​:P1​=x​:1.

Therefore,

x=4\sqrt{x} = 4x​=4 x=16.x = 16.x=16.
  1. Comparison with stored answer

Derived answer is 161616, which matches the stored correct answer.

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