Current Electricity Part 3

Drift velocity $= \left(\; \frac{e \tau}{m}\right) E$$v_d = \left(\; \frac{e \tau}{m}\right)\left(\; \frac{\Delta V}{\ell}\right)$$\Delta V =$ Potential difference applied across the wireAs temperature increases, relaxation time decreases, hence $V_d$ decreases.As per formula, $V_d \propto \frac{1}{\ell}$$v_d = \frac{I}{\text{neA}}$, as it is not mentioned that current is at steady state neither it is mentioned that $n$ is constant for given conductor. So it can't be said that $v_d$ is inversely proportional to $A$. $I = n e A v_d = \frac{V}{R} = \frac{V}{\rho \ell} A$$v_d = \frac{V}{\rho \ell \text{ne}} \;\; \left(E = \frac{V}{\ell}\right)$$v_d = \frac{eE \tau}{m}$$\tau$ decrease with temperature increase.First and fourth statements are correct.