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Current Electricity question

2022 · 26 Jun · Shift 1 · Q53
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  5. /2022 · 26 Jun · Shift 1 · Q53

Current Electricity question

2022 · 26 Jun · Shift 1 · Q53

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
An aluminium wire is stretched to make its length, 0.4% larger. The percentage change in resistance is :
  1. A
    0.4%
  2. B
    0.2%
  3. C
    0.8%
  4. D
    0.6%
View written solutionFree

Correct answer: C

  1. Resistance formula

For a wire, R=ρLAR = \rho \frac{L}{A}R=ρAL​ where ρ\rhoρ is resistivity, LLL is length, and AAA is cross-sectional area.

  1. Effect of stretching

When the wire is stretched, its volume remains constant: AL=constantAL = \text{constant}AL=constant So if length increases, area decreases accordingly.

Given the length increases by 0.4%0.4\%0.4%, ΔLL=0.4%=0.004\frac{\Delta L}{L} = 0.4\% = 0.004LΔL​=0.4%=0.004

Let the new length be L′=1.004LL' = 1.004LL′=1.004L

Since volume is constant, A′L′=ALA'L' = ALA′L′=AL A′=ALL′=A1.004A' = \frac{AL}{L'} = \frac{A}{1.004}A′=L′AL​=1.004A​

  1. New resistance

R′=ρL′A′=ρ1.004LA/1.004=ρLA(1.004)2R' = \rho \frac{L'}{A'} = \rho \frac{1.004L}{A/1.004} = \rho \frac{L}{A}(1.004)^2R′=ρA′L′​=ρA/1.0041.004L​=ρAL​(1.004)2 Thus, R′=R(1.004)2R' = R(1.004)^2R′=R(1.004)2

Now, (1.004)2=1.008016(1.004)^2 = 1.008016(1.004)2=1.008016

So percentage increase in resistance is R′−RR×100=(1.008016−1)×100\frac{R' - R}{R} \times 100 = (1.008016 - 1)\times 100RR′−R​×100=(1.008016−1)×100 =0.8016%= 0.8016\%=0.8016%

Approximately, 0.8%0.8\%0.8%

  1. Option check
  • A: 0.4%0.4\%0.4% ❌
  • B: 0.2%0.2\%0.2% ❌
  • C: 0.8%0.8\%0.8% ✅
  • D: 0.6%0.6\%0.6% ❌

Therefore, the correct answer is C.

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