\(v _{ d }=\left(\frac{ e \tau}{ m }\right)\left(\frac{\Delta V }{\ell}\right)\)

\(\Delta V=\) Potential difference applied across the wire

As 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 \(A\).

\(I=n e A v_{d}=\frac{V}{R}=\frac{V}{\rho \ell} A\)

\(v _{ d }=\frac{ V }{\rho \ell \text { ne }} \quad\left( E =\frac{ V }{\ell}\right)\)

\(v _{ d }=\frac{ eE \tau}{ m }\)

\(\tau\) decrease with temperature increase.

First and fourth statements are correct.