Consider a galvanometer shunted with $5\, \Omega$ resistance and $2\, \%$ of the total current passes through it. What is the resistance of the given galvanometer? (In $\Omega$)

$A$ 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 $\Omega$)

The value of the shunt resistance that allows $10 \%$ of the main current through the galvanometer of $99 \Omega$ is (in $\Omega$)

$A$ millivoltmeter of $25 \, mV$ range is to be converted into an ammeter of $25 \, A$ range. The value (in $\Omega$) of the necessary shunt will be:

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

$A$ moving coil galvanometer has a resistance of $50\,\Omega$ and gives full-scale deflection for $10\,mA$. How could it be converted into an ammeter with a full-scale deflection for $1\,A$?