The resistance of a galvanometer coil is $R$. What is the shunt resistance required to convert it into an ammeter of range $4$ times?

The resistance of a galvanometer is $90\, \Omega$. If only $10\%$ of the main current is to flow through the galvanometer,how should a resistor be connected and what should be its value?

$A$ maximum current of $0.5 \ mA$ can pass through a galvanometer of resistance $15 \ \Omega$. The resistance to be connected in series to the galvanometer to convert it into a voltmeter of range $0-10 \ V$ is (in $\Omega$)

In a moving coil galvanometer,to make the field radial,what is done?

$A$ 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 $\Omega$)

If only $\frac{1}{51}$ of the main current is to be passed through a galvanometer,then the shunt required is $R_1$. If only $\frac{1}{11}$ of the main voltage is to be developed across the galvanometer,then the resistance required is $R_2$. Find the ratio $\frac{R_2}{R_1}$.