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: