When the current $i$ is flowing through a conductor,the drift velocity is $v$. If $2i$ current is flowed through the same metal but having double the area of cross-section,then the drift velocity will be

$A$ cylindrical conductor of length $2 \ m$ and area of cross-section $0.2 \ mm^{2}$ carries an electric current of $1.6 \ A$ when its ends are connected to a $2 \ V$ battery. Mobility of electrons in the conductor is $\alpha \times 10^{-3} \ m^{2}/V \cdot s$. The value of $\alpha$ is: (electron concentration $n = 5 \times 10^{28} \ m^{-3}$ and electron charge $e = 1.6 \times 10^{-19} \ C$)

$A$ cylindrical wire of radius $0.5 \, mm$ and conductivity $5 \times 10^{7} \, S/m$ is subjected to an electric field of $10 \, mV/m$. The expected value of current in the wire will be $x^{3} \, \pi \, mA$. The value of $x$ is ......

$A$ copper wire of length $1 \,m$ and uniform cross-sectional area $5 \times 10^{-7} \,m^{2}$ carries a current of $1 \,A$. Assuming that there are $8 \times 10^{28}$ free electrons per $m^{3}$ in copper,how long will an electron take to drift from one end of the wire to the other?

When $5\ V$ potential difference is applied across a wire of length $0.1\ m$,the drift speed of electrons is $2.5 \times 10^{-4} \ m/s$. If the electron density in the wire is $8 \times 10^{28} \ m^{-3}$,the resistivity of the material is close to:

Derive the equation of mobility in terms of relaxation time $(\tau)$. Write its unit.