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JEE Main 2006 (AIEEE) · Ohm's Law, Resistivity & Temperature

Ohm's Law, Resistivity & Temperature question from JEE Main 2006 (AIEEE)

Q67JEE Main 2006 · AIEEEMedium

Ohm's Law, Resistivity & Temperature

The resistance of bulb filament is 100 Ω100\ \Omega at a temperature of 100∘C.100^\circ\text{C}. If its temperature coefficient of resistance be 0.0050.005 per ∘C^\circ\text{C}, its resistance will become 200 Ω200\ \Omega at a temperature of

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Answer: (b)

  1. Rt=R0(1+αt)R_t = R_0(1+\alpha t) with R0R_0 at 0∘0^\circC.
  2. 100=R0(1+0.005×100)=1.5R0⇒R0=200/3 Ω100 = R_0(1 + 0.005\times100) = 1.5R_0 \Rightarrow R_0 = 200/3\ \Omega
  3. 200=2003(1+0.005t)⇒1+0.005t=3⇒t=400∘200 = \tfrac{200}{3}(1 + 0.005t) \Rightarrow 1 + 0.005t = 3 \Rightarrow t = 400^\circC
  4. The answer is (b).
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