Resistance of a Galvanometer coil is \(8~\Omega\) and \(2~\Omega\) shunt resistance is connected with it. If main current is \(1\) A then the current flow through \(2~\Omega\) resistance will be:
1. \(0.2\) A
2. \(0.8\) A
3. \(0.1\) A
4. \(0.4\) A

A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a permanent magnet such that $\overrightarrow{\mathrm{B}}$ is in the plane of the coil. If due to a current i in the triangle, a torque τ acts on it, the side l of the triangle will be:

1. $\frac{2}{\sqrt{3}}\left(\frac{\tau }{\mathrm{Bi}}\right)$ $$

2. $\frac{1}{\sqrt{3}}\frac{\tau }{\mathrm{Bi}}$

3. $2{\left(\frac{\tau }{\sqrt{3}\mathrm{Bi}}\right)}^{\frac{1}{2}}$

4. $\frac{2}{\sqrt{3}}{\left(\frac{\tau }{\mathrm{Bi}}\right)}^{\frac{1}{2}}$

Two long parallel wires are at a distance of \(1\) m. If both of them carry one ampere of current in the same direction, then the force of attraction on the unit length of the wires will be:
1. \(2\times10^{-7}\) N/m
2. \(4\times10^{-7}\) N/m
3. \(8\times10^{-7}\) N/m
4. \(10^{-7}\) N/m

What properties will a galvanometer that is acting as a voltmeter have?

A current-carrying coil (I = 5A, R = 10 cm) has 50 turns. The magnetic field at its centre will be:
1.  1.57 mT
2.  3.14 mT
3.  1 mT
4.  2 mT

A galvanometer of \(50~\Omega\) resistance has \(25\) divisions. A current of \(4\times 10^{-4}~\text{A}\) gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of \(25~\text{V}\), it should be connected with a resistance of:

Two identically charged particles A and B initially at rest, are accelerated by a common potential difference V. They enter into a transverse uniform magnetic field B. If they describe a circular path of radii  $r_1$ $and$ $r_2$ respectively, then their mass ratio is:

1.  ${\left(\frac{r_1}{r_2}\right)}^2$

2. ${\left(\frac{r_2}{r_1}\right)}^2$

3. $\left(\frac{r_1}{r_2}\right)$

4. $\left(\frac{r_2}{r_1}\right)$

To convert a galvanometer into a voltmeter one should connect a:

The magnetic field of a given length of wire for a single-turn coil at its centre is 'B'. Its value for two turns coil for the same wire will be:

1. $\frac{\mathrm{B}}{4}$

2. $\frac{\mathrm{B}}{2}$

3. $4\mathrm{B}$

4. $2\mathrm{B}$

If a charge '\(q\)' moves with velocity \(\mathrm{v}\), in a region where electric field (\(\mathrm{E}\)) and magnetic field (\(\mathrm{B}\)) both exist, then force on it is: 
1. \(q(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})\)

2. \(q \overrightarrow{\mathrm{E}}+\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})\)

3. \( q \overrightarrow{\mathrm{E}}+q(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{v}})\)

4. \(\mathrm{qB}+\mathrm{q}(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{v}})\)