$A$ 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:

The resistance of an ammeter is $13\, \Omega$ and its scale is graduated for a current up to $100\, A$. After an additional shunt has been connected to this ammeter,it becomes possible to measure currents up to $750\, A$ by this meter. The value of the shunt resistance is:

In an experiment to determine the figure of merit of a galvanometer by the half-deflection method,a student constructed the following circuit. He unplugged a resistance of $5200 \ \Omega$ in $R$. When $K_1$ is closed and $K_2$ is open,the deflection observed in the galvanometer is $26 \ \text{div}$. When $K_2$ is also closed and a resistance of $90 \ \Omega$ is removed in $S$,the deflection becomes $13 \ \text{div}$. The resistance of the galvanometer is nearly: (in $\Omega$)

$A$ galvanometer of resistance $50 \Omega$ is connected to a battery of $3 \text{ V}$ along with a resistance of $2950 \Omega$ in series. $A$ full-scale deflection of $30$ divisions is obtained in the galvanometer. In order to reduce this deflection to $20$ divisions,the resistance in series should be: (in $\Omega$)

$A$ galvanometer of resistance $40\,\Omega$ gives a deflection of $5\, \text{divisions}$ per $mA$. There are $50\, \text{divisions}$ on the scale. The maximum current that can pass through it when a shunt resistance of $2\,\Omega$ is connected is ................ $mA$.

How can we convert a galvanometer into an ammeter?