$A$ galvanometer has a resistance of $100 \Omega$ and a current of $10 \text{ mA}$ produces full-scale deflection in it. The resistance to be connected to it in series,to convert it into a voltmeter of range $50 \text{ V}$,is: (in $\Omega$)

$A$ moving coil galvanometer, when shunted with a $2\text{ }\Omega$ resistance, gives a full-scale deflection for a current of $500\text{ mA}$. When a resistance of $470\text{ }\Omega$ is connected in series, it gives a full-scale deflection for a potential of $10\text{ V}$ applied across it. The value of the resistance of the galvanometer coil is . . . . . . $\Omega$.

$A$ galvanometer has a coil of resistance $200 \Omega$ with a full scale deflection at $20 \mu A$. The value of resistance to be added to use it as an ammeter of range $(0-20) mA$ is: (in $\Omega$)

$A$ galvanometer of resistance $50 \ \Omega$ giving full scale deflection for a current of $10 \ mA$ is to be changed into a voltmeter of range $100 \ V$. $A$ resistance of . . . . . . $\Omega$ has to be connected in series with the galvanometer.

$A$ galvanometer having coil resistance $10 \ \Omega$ shows a full scale deflection for a current of $3 \ mA$. For it to measure a current of $8 \ A$,the value of the shunt should be:

When a moving coil galvanometer $(MCG)$ is converted into a voltmeter,the series resistance is '$n$' times the resistance of the galvanometer. How many times the original voltage range of the $MCG$ is the voltmeter now capable of measuring?