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

$A$ galvanometer coil has a resistance of $12\,\Omega$ and the meter shows full-scale deflection for a current of $3\,mA$. To convert it into a voltmeter of range $0-18\,V$,what resistance should be added and how?

$A$ student connects a voltmeter,an ammeter,and a resistor as shown in the circuit. If the voltmeter reading is $20 \ V$ and the ammeter reading is $4 \ A$,then the resistance $R$ will be:

$A$ moving coil galvanometer has $48$ turns and the area of the coil is $4 \times 10^{-2} \, m^2$. If the magnetic field is $0.2 \, T$,then to increase the current sensitivity by $25\%$ without changing the area $(A)$ and the magnetic field $(B)$,the number of turns should become:

The ratios of the voltage sensitivities,resistances and areas of the coils of two moving coil galvanometers $A$ and $B$ are $4:3$,$3:4$ and $1:2$ respectively. If the number of turns of the coil of galvanometer $A$ is $200$,then the number of turns of the coil of galvanometer $B$ is (All other quantities remain same in both the cases)

In the diagram shown,the reading of the voltmeter is $20\, V$ and that of the ammeter is $4\, A$. The value of $R$ should be (Consider the given ammeter and voltmeter are not ideal).