If only $2\%$ of the main current is to be passed through a galvanometer of resistance $G$,then the resistance of the shunt will be

$A$ galvanometer coil has a resistance of $12\; \Omega$ and the meter shows full-scale deflection for a current of $3\; mA$. How will you convert the meter into a voltmeter of range $0$ to $18\; V$?

$A$ maximum current of $0.5 \ mA$ can pass through a galvanometer of resistance $15 \ \Omega$. The resistance to be connected in series to the galvanometer to convert it into a voltmeter of range $0-10 \ V$ is (in $\Omega$)

In an experiment to find the resistance of a galvanometer by the half-deflection method, a $5 V$ battery and a high resistance of $4.9 k\Omega$ are connected in the circuit. In the absence of any shunt resistance, the galvanometer reads $20$ divisions when current flows in the circuit. To reduce the deflection by half, the value of the shunt resistance used is $98 \Omega$. The figure of merit of the galvanometer is given as $..... \mu A / \text{division}$.

The phosphor bronze strip is used in a moving coil galvanometer because

$A$ galvanometer of resistance $50 \ \Omega$ is connected to a battery of $8 \ V$ along with a resistance of $3950 \ \Omega$ in series. $A$ full-scale deflection of $30$ divisions is obtained in the galvanometer. In order to reduce this deflection to $15$ divisions,the resistance in series should be . . . . . . $\Omega$.