In a circuit, \(5\) C of charge is passed through a battery in a given time. The plates of the battery are maintained at a potential difference of \(12\) V. The work done by the battery is:

A bullet of mass \(2~\text {gm}\) has a charge of \(2~\mu\text{C}.\) Through what potential difference must it be accelerated, starting from rest, to acquire a speed of \(10~\text{m/s}?\)
1. \(50~\text {kV}\)
2. \(5~\text {V}\)
3. \(50~\text {V}\)
4. \(5~\text {kV}\)

Four electric charges \(+ q,\) \(+ q,\) \(- q\) and \(- q\) are placed at the corners of a square of side \(2L\) (see figure). The electric potential at the point \(A\), mid-way between the two charges \(+ q\) and \(+ q\) is:
              
1. \(\frac{1}{4 \pi\varepsilon_{0}} \frac{2 q}{L} \left(1 + \frac{1}{\sqrt{5}}\right)\)
2. \(\frac{1}{4 \pi\varepsilon_{0}} \frac{2 q}{L} \left(1 - \frac{1}{\sqrt{5}}\right)\)
3. zero
4. \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q}{L} \left(1 + \sqrt{5}\right)\)

Twenty seven drops of same size are charged at \(220~\text{V}\) each. They combine to form a bigger drop. Calculate the potential of the bigger drop:
1. \(1520~\text{V}\)
2. \(1980~\text{V}\)
3. \(660~\text{V}\)
4. \(1320~\text{V}\)

If \(50~\text{J}\) of work must be done to move an electric charge of \(2~\text{C}\) from a point where the potential is \(-10\) volts to another point where the potential is \(\mathrm{V}\) volts, then the value of \(\mathrm{V}\) is:
1. \(5\) volts
2. \(-15\) volts
3. \(+15\) volts
4. \(+10\) volts

A short electric dipole has a dipole moment of \(16 \times 10^{-9} ~\text{C-}\text{m}\). The electric potential due to the dipole at a point at a distance of \(0.6~\text{m}\) from the centre of the dipole situated on a line making an angle of \(60^{\circ}\) with the dipole axis is: \(\left( \dfrac{1}{4\pi \varepsilon_0}= 9\times 10^{9}~\text{N-m}^2/\text{C}^2\right)\)
1. \(200~\text{V}\)
2. \(400~\text{V}\)
3. zero
4. \(50~\text{V}\)

A charge \(+q\) is fixed at each of the points $x=x_0,$ $x=3x_0,$ $x=5x_0$ ..... infinite, on the \(x\)-axis, and a charge \(-q\) is fixed at each of the points $x=2x_0,$ $x=4x_0,x=6x_0$,..... infinite. Here \(x_0\) is a positive constant. Take the electric potential at a point due to a charge \(Q\) at a distance \(r\) from it to be \(\frac{Q}{4\pi \varepsilon_0 r}\). Then, the potential at the origin due to the above system of charges is:
1. \(0\)
2. \(\frac{q}{8 \pi \varepsilon_{0} x_{0} \mathrm{ln} 2}\)
3. \(\infty\)
4. \(\frac{q \mathrm{ln} 2}{4 \pi \varepsilon_{0} x_{0}}\)