How can we convert a galvanometer into a voltmeter?

Two moving coil galvanometers $A$ and $B$ having identical springs are placed in magnetic fields of $0.25 \ T$ and $0.5 \ T$ respectively. If the number of turns in $A$ and $B$ are respectively $36$ and $48$,and the areas of the coils $A$ and $B$ are $2.4 \times 10^{-3} \ m^2$ and $4.8 \times 10^{-3} \ m^2$ respectively,then the ratio of the current sensitivities of the galvanometers $A$ and $B$ is

To verify Ohm's law,a student is provided with a test resistor $R_T$,a high resistance $R_1$,a small resistance $R_2$,two identical galvanometers $G_1$ and $G_2$,and a variable voltage source $V$. The correct circuit to carry out the experiment is

$A$ galvanometer has a coil of resistance $100 \Omega$ showing a full scale deflection at $50 \mu A$. The resistance that should be added to use it as an ammeter of range $10 \text{ mA}$ is

Explain the equation of a shunt.

The resistance of a $1\, A$ ammeter is $0.018\,\Omega$. To convert it into a $10\, A$ ammeter,the shunt resistance required will be: