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 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 resistance of a wire is \(R\) ohm. If it is melted and stretched to \(n\) times its original length, its new resistance will be:
1. \(nR\)
2. \(\frac{R}{n}\)
3. \(n^2R\)
4. \(\frac{R}{n^2}\)

Two solid conductors are made up of the same material and have the same length and the same resistance. One of them has a circular cross-section of area $A_1$ and the other one has a square cross-section of area $A_2$. The ratio $A_1/A_2$ is:

The dependence of resistivity \((\rho)\) on the temperature \((T)\) of a semiconductor is, roughly, represented by:

1.		2.
3.		4.

The equivalent resistance between \(A\) and \(B\) for the mesh shown in the figure is:

A potential divider is used to give outputs of \(2~\text{V}\) and \(3~\text{V}\) from a \(5~\text{V}\) source, as shown in the figure.

In the circuit shown in the figure, the effective resistance between \(A\) and \(B\) is:

1. \(2~\Omega\)
2. \(4~\Omega\)
3. \(6~\Omega\)
4. \(8~\Omega\)

The effective resistance between points \(P\) and \(Q\) of the electrical circuit shown in the figure is:
            
1. \(\frac{2 R r}{\left(R + r \right)}\)
2. \(\frac{8R\left(R + r\right)}{\left( 3 R + r\right)}\)
3. \(2r+4R\)
4. \(\frac{5R}{2}+2r\)