If a galvanometer is shunted by $(\frac{1}{n-1})^{th}$ of the value of its resistance, then the fraction of the total current passing through the galvanometer is

The resistance of an ideal voltmeter is . . . . . . .

In the experiment to determine the galvanometer resistance by the half-deflection method,the plot of $\frac{1}{\theta}$ vs the resistance $(R)$ of the resistance box is shown in the figure. The figure of merit of the galvanometer is .............. $\times 10^{-1} \text{ A/division}$. [The source has emf $E = 2 \text{ V}$]

$A$ moving coil galvanometer has $100$ 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 Ohm's needed to be connected in series with the coil will be

$A$ 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 $\Omega$)

When a shunt resistance of $4r$ is connected to a galvanometer,it becomes an ammeter that can measure $0.03 \, A$. When a shunt resistance of $r$ is connected to the same galvanometer,it becomes an ammeter that can measure $0.06 \, A$. What is the current capacity $(i_g)$ of the galvanometer in $A$?