$A$ galvanometer has a resistance of $7\,\Omega$ and gives a full-scale deflection for a current of $1.0\,A$. How will you convert it into a voltmeter of range $10\,V$?

In an experiment to determine the figure of merit of a galvanometer by the half-deflection method,a student constructed the following circuit. He unplugged a resistance of $5200 \ \Omega$ in $R$. When $K_1$ is closed and $K_2$ is open,the deflection observed in the galvanometer is $26 \ \text{div}$. When $K_2$ is also closed and a resistance of $90 \ \Omega$ is removed in $S$,the deflection becomes $13 \ \text{div}$. The resistance of the galvanometer is nearly: (in $\Omega$)

$A$ voltmeter has a resistance of $G \, \Omega$ and a range of $V \, \text{volts}$. The value of the resistance that must be connected in series to convert it into a voltmeter of range $nV \, \text{volts}$ is:

To convert a galvanometer into an ammeter,one needs to connect a

To make the magnetic field radial in a moving coil galvanometer:

$A$ galvanometer having a coil resistance $100 \; \Omega$ gives a full-scale deflection when a current of $1 \; mA$ is passed through it. What is the value of the resistance in $k\Omega$ which can convert this galvanometer into a voltmeter giving full-scale deflection for a potential difference of $10 \; V$?