Match the LIST-I with LIST-II (Symbols have their usual meaning)LIST-…

The correct answer is Option (1) → A-IV, B-III, C-II, D-I

All four expressions come from the drift model: in a field E each free electron gains an acceleration \( eE/m \) for an average time \( \tau \) (the relaxation time) between collisions.

(A) Drift Velocity — this matches (IV) \( \tau eE/m \). The electron gains velocity at the rate \( eE/m \) for a time \( \tau \).

    \( v_d = a\tau = eE\tau/m \)

(B) Mobility — this matches (III) \( e\tau/m \). Mobility is the drift velocity per unit field.

    \( \mu = v_d/E = e\tau/m \)

(C) Conductivity — this matches (II) \( l/(RA) \). Conductivity is the reciprocal of resistivity, and \( R = \rho l/A \) gives \( \rho = RA/l \).

    \( \displaystyle \sigma = \frac{1}{\rho} = \frac{l}{RA} \)

(D) Resistivity — this matches (I) \( m/(ne^{2}\tau) \). Putting \( v_d \) into \( j = nev_d \) gives \( j = (ne^{2}\tau/m)E = \sigma E \).

    \( \displaystyle \rho = \frac{1}{\sigma} = \frac{m}{ne^{2}\tau} \)

NCERT (2026-27):
Drift of Electrons and the Origin of Resistivity – “… Thus, averaging Eq. (3.16) over the N-electrons at any given time t gives us for the average velocity v_d

    \( \displaystyle \mathbf{v}_d \equiv \left(\mathbf{v}_i\right)_{\text{average}} = -\frac{e\mathbf{E}}{m}\,\tau \qquad (3.17) \)

This is the phenomenon of drift and the velocity v_d in Eq. (3.17) is called the drift velocity.”

Mobility – “… An important quantity is the mobility \( \mu \) defined as the magnitude of the drift velocity per unit electric field: … Hence,

    \( \displaystyle \mu = \frac{v_d}{E} = \frac{e\tau}{m} \qquad (3.25) \)”

Ohm’s Law – “… \( R = \rho l/A \) (3.9) … or, \( \mathbf{j} = \sigma\mathbf{E} \) (3.13) where \( \sigma \equiv 1/\rho \) is called the conductivity.”

Temperature Dependence of Resistivity – “… From this equation, resistivity of a material is given by

    \( \displaystyle \rho = \frac{1}{\sigma} = \frac{m}{n e^{2} \tau} \qquad (3.27) \)”