$A$ milliammeter of range $10 \, mA$ has a coil of resistance $1 \, \Omega$. To use it as a voltmeter of range $10 \, V$,the resistance that must be connected in series with it will be ............. $\Omega$.

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 of resistance $240 \Omega$ allows only $4 \%$ of the main current to pass through it after connecting a shunt resistance. The value of the shunt resistance is (in $\Omega$)

$A$ milliammeter of resistance $40 \Omega$ has a range $0-30 \text{ mA}$. What will be the resistance used in series to convert it into a voltmeter of range $0-15 \text{ V}$ (in $\Omega$)?

$A$ galvanometer of resistance $100\,\Omega$ has $50$ divisions on its scale and a sensitivity of $20\,\mu A/\text{division}$. It is to be converted into a voltmeter with three ranges: $0-2\,V$,$0-10\,V$,and $0-20\,V$. The appropriate circuit to do so is:

When a resistance of $200 \Omega$ is connected in series with a galvanometer of resistance $G$,its range is $V$. To triple its range,a resistance of $2000 \Omega$ is connected in series. The value of $G$ is (in $\Omega$)