To convert a galvanometer into an ammeter,we connect . . . . . . .

$A$ galvanometer has a deflection of $25$ divisions for a current of $0.1 \ A$ flowing through it. Then the current sensitivity of the galvanometer is (in $div/A$)

$A$ galvanometer having a resistance of $9 \, \Omega$ is shunted by a wire of resistance $2 \, \Omega$. If the total current is $1 \, A$, the part of it passing through the shunt will be ............ $A$.

An ammeter with a resistance of $2 \,\Omega$ can measure up to $100 \,mA$. What resistance (in $\Omega$) must be connected in series to make it measure up to $5 \,V$?

Only $4 \%$ of the total current in the circuit passes through a galvanometer. If the resistance of the galvanometer is $G$,then the shunt resistance connected to the galvanometer is:

$A$ galvanometer having $30$ divisions has a current sensitivity of $0.0625 \frac{\text{div}}{\mu A}$. If it is converted into a voltmeter to read a maximum of $6 \text{ V}$,then the resistance of that voltmeter is: