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

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

Q66JEE Main 2006 · AIEEEEasy

Ohm's Law, Resistivity & Temperature

A material 'B' has twice the specific resistance of 'A'. A circular wire made of 'B' has twice the diameter of a wire made of 'A'. Then for the two wires to have the same resistance, the ratio ℓB/ℓA\ell_B/\ell_A of their respective lengths must be

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Answer: (a)

  1. R=ρlπd2/4⇒l=Rπd24ρR = \dfrac{\rho l}{\pi d^2/4} \Rightarrow l = \dfrac{R \pi d^2}{4\rho}, so for the same RR, l∝d2ρl \propto \dfrac{d^2}{\rho}.
  2. lBlA=dB2dA2⋅ρAρB=4×12\dfrac{l_B}{l_A} = \dfrac{d_B^2}{d_A^2} \cdot \dfrac{\rho_A}{\rho_B} = 4 \times \dfrac{1}{2}
  3. lBlA=2\dfrac{l_B}{l_A} = 2
  4. So the answer is (a) 22.
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