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 relation i= (3+2t) A. The amount of charge passing a cross section of the wire in the time interval t=0 to t=4.0 sec would be: (where t is time in seconds)

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 \(7.5\times10^{-4}~\mathrm{ms^{-1}}\) in an electric field of \(3\times10^{-10}\) Vm^-1, has mobility in ${\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$ of:
1. $2.5\times {10}^6$
2. $2.5\times {10}^{-6}$
3. $2.25\times {10}^{-15}$
4. $2.25\times {10}^{15}$

Drift velocity v_d varies with the intensity of electric field as per the relation: 

1. $v_d\propto E$

2. $v_d\propto \frac{1}{E}$

3. v_d = 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:
1.  \(v\)
2. \(\frac{v}{2}\)$$
3. \(\frac{v}{4}\)$$
4. \(\frac{v}{8}\)

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