$A$ 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?

$A$ galvanometer acting as a voltmeter should have . . . . . . .

Two galvanometers $A$ and $B$ require $3\,mA$ and $5\,mA$ respectively to produce the same deflection of $10$ divisions. Then:

$A$ galvanometer may be converted into an ammeter or a voltmeter. In which of the following cases will the resistance of the device be the largest? (Assume the maximum range of the galvanometer is $1 \, mA$)

$A$ galvanometer of resistance $50 \Omega$ is connected to a battery of $3 \text{ V}$ along with a resistance of $2950 \Omega$ in series,showing a full-scale deflection of $30$ divisions. The additional series resistance required to reduce the deflection to $20$ divisions is: (in $\Omega$)

$A$ 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.