If the potential difference across ends of a metallic wire is doubled, the drift velocity of charge carriers will become:
1. double 
2. half
3. four times
4. one-fourth 

The current in a wire varies with time according to the equation $I=(4+2t),$ where $I$ is in ampere and $t$ is in seconds. The quantity of charge which has passed through a cross-section of the wire during the time $t=2$ s to $t=6$ s will be:

A charged particle having drift velocity of $7.5\times10^{-4}~\text{ms}^{-1}$ in an electric field of $3\times10^{-10}~\text{Vm}^{-1},$ has mobility of: 
1. $2.5\times 10^{6}~\text{m}^2\text{V}^{-1}\text{s}^{-1}$
2. $2.5\times 10^{-6}~\text{m}^2\text{V}^{-1}\text{s}^{-1}$
3. $2.25\times 10^{-15}~\text{m}^2\text{V}^{-1}\text{s}^{-1}$
4. $2.25\times 10^{15}~\text{m}^2\text{V}^{-1}\text{s}^{-1}$

Drift velocity $v_d$ varies with the intensity of the electric field as per the relation:
1. $v_{d} \propto E$
2. $v_{d} \propto \frac{1}{E}$
3. $v_{d}= \text{constant}$
4. $v_{d} \propto E^2$

The drift velocity of free electrons in a conductor is $v$ when a current $i$ is flowing in it. If both the radius and current are doubled, then the drift velocity will be:

A current passes through a wire of variable cross-section in steady-state as shown. Then incorrect statement is: