When an additional resistance $1980 \ \Omega$ is connected in series with a voltmeter,each scale division has $100$ times larger value. The resistance of the voltmeter is (in $\Omega$)

$A$ galvanometer has resistance $G \ \Omega$ and $I_g$ is the current flowing through it which produces full-scale deflection. $S_1$ is the value of the shunt which converts it into an ammeter of range $0$ to $3I$,and $S_2$ is the shunt value which converts it into an ammeter of range $0$ to $4I$. The ratio $S_2:S_1$ is:

$A$ galvanometer of resistance $G$ has a voltage range $V_g$. The resistance required to convert it to read voltage up to $V$ is

$A$ milliammeter of resistance $40 \Omega$ has a range $0-30 \text{ mA}$. What will be the resistance used in series to convert it into a voltmeter of range $0-15 \text{ V}$ (in $\Omega$)?

If only $2\%$ of the main current is to be passed through a galvanometer of resistance $G$,then the resistance of the shunt will be

What is a galvanometer? Give its applications.