$A$ current $I$ is passing through a wire having two sections $P$ and $Q$ of uniform diameters $d$ and $d/2$ respectively. If the mean drift velocity of electrons in sections $P$ and $Q$ is denoted by $v_P$ and $v_Q$ respectively,then

$A$ potential difference $V$ is applied across the ends of a metallic wire of length $l$. The drift velocity will be doubled if:

Assertion: The current density $\vec J$ at any point in an ohmic resistor is in the direction of the electric field $\vec E$ at that point.
Reason: $A$ point charge when released from rest in a region having only an electrostatic field always moves along electric lines of force.

$A$ current $I$ flows through a uniform wire of diameter $d$,when the mean drift velocity is $v_d$. The same current will flow through a wire of diameter $d/2$ made of the same material,if the mean drift velocity of the electrons is

$A$ copper wire of length $1 \ m$ and radius $1 \ mm$ is connected in series with an iron wire of length $2 \ m$ and radius $3 \ mm$. If a current flows through both wires,the ratio of the current density in the copper wire to that in the iron wire 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 $e = 1.6 \times 10^{-19} \ C$)