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JEE Main 2019 (9 Apr, Shift 2) · Electric Current, Drift Velocity & Current Density

Electric Current, Drift Velocity & Current Density question from JEE Main 2019 (9 Apr, Shift 2)

Q28JEE Main 2019 · 9 APR · SHIFT 2Medium

Electric Current, Drift Velocity & Current Density

In a conductor, if the number of conduction electrons per unit volume is 8.5 × 1028 m–3 and mean free time is 25ƒs (femto second), it's approximate resistivity is :-

(me = 9.1 × 10–31 kg)

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Answer: (a)

  1. ρ=mne2τ\rho = \dfrac{m}{ne^{2}\tau}
  2. ρ=9.1×10−318.5×1028×(1.6×10−19)2×25×10−15=9.1×10−315.44×10−23\rho = \dfrac{9.1 \times 10^{-31}}{8.5 \times 10^{28} \times (1.6 \times 10^{-19})^{2} \times 25 \times 10^{-15}} = \dfrac{9.1 \times 10^{-31}}{5.44 \times 10^{-23}}
  3. ρ≈1.7×10−8 Ωm∼10−8 Ωm\rho \approx 1.7 \times 10^{-8}\ \Omega\text{m} \sim 10^{-8}\ \Omega\text{m}
  4. So the answer is (a).
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