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

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:

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

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

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

4. \(q\vec{B}+{q}(\vec{E} \times \vec{v})\)