$A$ current of $2 \,A$ is passing through a metal wire of cross-sectional area $2 \times 10^{-6} \,m^{2}$. If the number density of free electrons in the wire is $5 \times 10^{26} \,m^{-3}$, the drift speed of electrons is (Given, $e = 1.6 \times 10^{-19} \,C$)

$A$. The drift velocity of electrons decreases with the increase in the temperature of a conductor.
$B$. The drift velocity is inversely proportional to the area of cross-section of a given conductor.
$C$. The drift velocity does not depend on the applied potential difference to the conductor.
$D$. The drift velocity of an electron is inversely proportional to the length of the conductor.
$E$. The drift velocity increases with the increase in the temperature of a conductor.
Choose the correct answer from the options given below:

Which particles are responsible for the conductivity of metals?

$A$ metal has $9 \times 10^{28}$ conduction electrons per $m^3$ and its resistivity is $1 \times 10^{-8} \Omega \cdot m$. If the drift speed of an electron in the metal is $1.6 \times 10^6 \ m/s$,then its mean free path is (mass of electron $= 9 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$). (in $nm$)

The drift velocity of electrons in a silver wire with a cross-sectional area of $3.14 \times 10^{-6} \, m^2$ carrying a current of $20 \, A$ is. Given the atomic weight of $Ag = 108$ and the density of silver $= 10.5 \times 10^3 \, kg/m^3$,the drift velocity is $.......... \times 10^{-4} \, m/s$.

The drift velocity of an electron is $v_d$ in a conductor of area of cross-section $A$ and carries a current $I$. Now,the area of cross-section and current flowing through the conductor are doubled,then the new drift velocity of the electron is . . . . . . .