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:

1.	\(60\) C	2.	\(24\) C
3.	\(48\) C	4.	\(30\) C

The Wheatstone bridge shown in the figure below is balanced when the uniform slide wire AB is divided as shown. Value of the resistance X is:
        

1. 3 $\mathrm{\Omega }$

2. 4 $\mathrm{\Omega }$

3. 2 $\mathrm{\Omega }$

4. 7 $\mathrm{\Omega }$

The total power dissipated in watts in the circuit shown below is:

A wire of resistance \(12~ \Omega \text{m}^{-1}\)${\mathrm{}}^{}$ is bent to form a complete circle of radius \(10~\text{cm}\). The resistance between its two diametrically opposite points, \(A\) and \(B\) as shown in the figure, is:
        
1. \(0.6\pi~\Omega\)$$
2. \(3\pi ~\Omega\)$$
3. \(61 \pi~ \Omega\)$$
4. \(6\pi~\Omega\)$$

Given below two statements:

If the voltage across a bulb rated \((220~\text{V}\text-100~\text{W})\) drops by \(2.5\%\) of its rated value, the percentage of the rated value by which the power would decrease is:
1. \(20\%\)
2. \(2.5\%\)
3. \(5\%\)
4. \(10\%\)

A potentiometer wire of length \(L\) and a resistance \(r\) are connected in series with a battery of EMF \(E_{0 }\) and resistance \(r_{1}\). An unknown EMF is balanced at a length l of the potentiometer wire. The EMF \(E\) will be given by:
1. \(\frac{L E_{0} r}{l r_{1}}\)
2. \(\frac{E_{0} r}{\left(\right. r + r_{1} \left.\right)} \cdot \frac{l}{L}\)
3. \(\frac{E_{0} l}{L}\)
4. \(\frac{L E_{0} r}{\left(\right. r + r_{1} \left.\right) l}\)