$A$ $100\, \Omega$ galvanometer gives full-scale deflection at $10\, mA$. How much shunt is required to read $100\, mA$?

$A$ current of $10^{-3} \ A$ is flowing through a resistance of $1000 \ \Omega$. To measure the correct potential difference across this resistance,a voltmeter must be used whose resistance should be:

$A$ galvanometer coil has a resistance $80 \Omega$ and current for full-scale deflection is $10 \text{ mA}$. $A$ resistance of $920 \Omega$ is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is $0.2 \text{ V}$,the number of divisions on the scale is:

$A$ galvanometer of resistance $50 \Omega$ is connected to a battery of $3 \text{ V}$ along with a resistance of $2950 \Omega$ in series. $A$ full-scale deflection of $30$ divisions is obtained in the galvanometer. In order to reduce this deflection to $20$ divisions,the resistance in series should be: (in $\Omega$)

$A$ galvanometer of resistance $G$ is shunted by a resistance $S$. To keep the main current in the circuit unchanged,the resistance to be put in series with the galvanometer is

In an ammeter,$5 \%$ of the main current passes through the galvanometer. If the resistance of the galvanometer is $G$,the resistance of the ammeter will be: