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Current Electricity question

2022 · 28 Jun · Shift 2 · Q60
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Current Electricity question

2022 · 28 Jun · Shift 2 · Q60

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
Resistance of the wire is measured as 2 Ω\OmegaΩ and 3 Ω\OmegaΩ at 10 ∘^\circ∘ C and 30 ∘^\circ∘ C respectively. Temperature co-efficient of resistance of the material of the wire is :
  1. A
    0.033 ∘^\circ∘ C −-− 1
  2. B
    −-− 0.033 ∘^\circ∘ C −-− 1
  3. C
    0.011 ∘^\circ∘ C −-− 1
  4. D
    0.055 ∘^\circ∘ C −-− 1
View written solutionFree

Correct answer: A

  1. Use the linear relation for resistance with temperature

For a conductor, Rt=R0(1+αt)R_t = R_0(1+\alpha t)Rt​=R0​(1+αt) where:

  • RtR_tRt​ is resistance at temperature ttt,
  • R0R_0R0​ is resistance at 0∘0^\circ0∘C,
  • α\alphaα is temperature coefficient of resistance.

Given: R10=2 Ω,R30=3 ΩR_{10} = 2\,\Omega, \qquad R_{30} = 3\,\OmegaR10​=2Ω,R30​=3Ω

So, 2=R0(1+10α)...(1)2 = R_0(1+10\alpha) \quad ...(1)2=R0​(1+10α)...(1) 3=R0(1+30α)...(2)3 = R_0(1+30\alpha) \quad ...(2)3=R0​(1+30α)...(2)

  1. Divide equation (2) by equation (1)

32=1+30α1+10α\frac{3}{2} = \frac{1+30\alpha}{1+10\alpha}23​=1+10α1+30α​

Cross-multiplying: 3(1+10α)=2(1+30α)3(1+10\alpha) = 2(1+30\alpha)3(1+10α)=2(1+30α)

3+30α=2+60α3 + 30\alpha = 2 + 60\alpha3+30α=2+60α

1=30α1 = 30\alpha1=30α

α=130=0.033 ∘C−1\alpha = \frac{1}{30} = 0.033\,^\circ \text{C}^{-1}α=301​=0.033∘C−1

  1. Match with the options
  • A: 0.033 ∘C−10.033\,^\circ \text{C}^{-1}0.033∘C−1 ✅
  • B: −0.033 ∘C−1-0.033\,^\circ \text{C}^{-1}−0.033∘C−1
  • C: 0.011 ∘C−10.011\,^\circ \text{C}^{-1}0.011∘C−1
  • D: 0.055 ∘C−10.055\,^\circ \text{C}^{-1}0.055∘C−1

Hence, the correct option is A.

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