$A$ galvanometer of resistance $50 \ \Omega$ giving full scale deflection for a current of $10 \ mA$ is to be changed into a voltmeter of range $100 \ V$. $A$ resistance of . . . . . . $\Omega$ has to be connected in series with the galvanometer.

$A$ moving coil galvanometer has $150$ equal divisions. Its current sensitivity is $10$ divisions per milliampere and voltage sensitivity is $2$ divisions per millivolt. In order that each division reads $1 \, V$,the resistance in $\Omega$ needed to be connected in series with the coil will be

$A$ galvanometer with $30$ divisions has a current sensitivity of $20 \, \mu A/division$. Its resistance is $25 \, \Omega$. How will you convert it into an ammeter to measure $1 \, A$ current?

Two galvanometers $A$ and $B$ require currents of $4 \ mA$ and $7 \ mA$,respectively,to produce the same deflection of $20$ divisions. If $S_{A}$ and $S_{B}$ are their sensitivities,respectively,then:

$A$ 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:

$A$ moving coil galvanometer is converted into an ammeter reading up to $0.03\,A$ by connecting a shunt of resistance $4r$ across it,and into an ammeter reading up to $0.06\,A$ when a shunt of resistance $r$ is connected across it. What is the maximum current which can be sent through this galvanometer if no shunt is used?