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Current Electricity question

2011 · Shift 0 · Q62
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Current Electricity question

2011 · Shift 0 · Q62

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
If a wire is stretched to make it 0.1%0.1\%0.1% longer, its resistance will:
  1. A
    increase by 0.2%0.2\%0.2%
  2. B
    decrease by 0.2%0.2\%0.2%
  3. C
    decrease by 0.05%0.05\%0.05%
  4. D
    increase by 0.05%0.05\%0.05%
View written solutionFree

Correct answer: A

  1. Use the formula for resistance

For a wire, R=ρLAR = \rho \frac{L}{A}R=ρAL​ where:

  • ρ\rhoρ = resistivity,
  • LLL = length,
  • AAA = cross-sectional area.

Assume stretching does not change resistivity and the volume remains constant.

  1. Apply constant volume condition

Since volume V=ALV = ALV=AL is constant, AL=constantAL = \text{constant}AL=constant So, A∝1LA \propto \frac{1}{L}A∝L1​

  1. Express resistance in terms of length only

Since R=ρLAR = \rho \frac{L}{A}R=ρAL​ and A∝1LA \propto \frac{1}{L}A∝L1​, we get R∝L2R \propto L^2R∝L2

  1. Find fractional change in resistance

If the wire is stretched to make it 0.1%0.1\%0.1% longer, then ΔLL=0.1%=0.001\frac{\Delta L}{L} = 0.1\% = 0.001LΔL​=0.1%=0.001

Because R∝L2R \propto L^2R∝L2, ΔRR=2ΔLL\frac{\Delta R}{R} = 2\frac{\Delta L}{L}RΔR​=2LΔL​ for small changes.

Thus, ΔRR=2(0.1%)=0.2%\frac{\Delta R}{R} = 2(0.1\%) = 0.2\%RΔR​=2(0.1%)=0.2%

So the resistance increases by 0.2%0.2\%0.2%.

  1. Check options
  • A: increase by 0.2%0.2\%0.2% ✅
  • B: decrease by 0.2%0.2\%0.2% ❌
  • C: decrease by 0.05%0.05\%0.05% ❌
  • D: increase by 0.05%0.05\%0.05% ❌

Therefore, the correct option is A.

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