Verified Solution • 2025–26 Session moderate

QJEEM15Phd13d: JEE Main 2015 Physics - Class 12 JEEM15Phd13d Solution: When potential difference is applied across a wire of length…

When 5V5V potential difference is applied across a wire of length 0.10.1 m,m, the drift speed of electrons is 2.5×10−4  ms−1.2.5 \times {10^{ - 4}}\,\,m{s^{ - 1}}. If the electron density in the wire is 8×1028  m−3,8 \times {10^{28}}\,\,{m^{ - 3}}, the resistivity of the material is close to :

Multiple Choice Options

  • A 1.6×10−6Ωm1.6 \times {10^{ - 6}}\Omega m
  • B 1.6×10−5Ωm1.6 \times {10^{ - 5}}\Omega m
  • C 1.6×10−8Ωm1.6 \times {10^{ - 8}}\Omega m
  • D 1.6×10−7Ωm1.6 \times {10^{ - 7}}\Omega m

Step-by-Step Verified Solution

V=IR=(neAvd)ρℓAV = IR = \left( {neA{v_d}} \right)\rho {\ell \over A}

∴\therefore ρ=VVdln⁡e\rho = {V \over {{V_d}\ln e}}

Here V=V= potential difference

l=l = length of wire

n=n= no. of electrons per unit volume of conductor.

e=e= no. of electrons

Placing the value of above parameters we get resistivity

ρ=58×1028×1.6×10−19×2.5×10−4×0.1\rho = {5 \over {8 \times {{10}^{28}} \times 1.6 \times {{10}^{ - 19}} \times 2.5 \times {{10}^{ - 4}} \times 0.1}}

=1.6×10−5Ωm= 1.6 \times {10^{ - 5}}\Omega m

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