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JEE Main 2023 (12 Apr, Shift 1) · Ohm's Law, Resistivity & Temperature

Ohm's Law, Resistivity & Temperature question from JEE Main 2023 (12 Apr, Shift 1)

Q19JEE Main 2023 · 12 APR · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature

The current flowing through a conductor connected across a source is 2 A2\text{ A} and 1.2 A\text{A} at 0∘C0^\circ\text{C} and 100∘C100^\circ\text{C} respectively. The current flowing through the conductor at 50∘C50^\circ\text{C} will be ___________ ×102 mA\times 10^{2}\text{ mA}.

Numerical answer type. Work it out, then open the answer.

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Answer: 15

  1. I=V/RI = V/R, so R0=V/2R_0 = V/2 and R100=V/1.2R_{100} = V/1.2.
  2. R100=R0(1+100α)⇒1+100α=2/1.2=5/3⇒α=1/150R_{100} = R_0(1+100\alpha) \Rightarrow 1+100\alpha = 2/1.2 = 5/3 \Rightarrow \alpha = 1/150.
  3. R50=R0(1+50/150)=43⋅V2=2V3R_{50} = R_0(1+50/150) = \tfrac43\cdot\tfrac{V}{2} = \tfrac{2V}{3}
  4. I50=1.5 A=1500 mA=15×102I_{50} = 1.5\ \text{A} = 1500\ \text{mA} = 15\times10^2 mA, so the answer is 15.
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