To convert a $800\, mV$ range millivoltmeter of resistance $40\,\Omega$ into an ammeter of $100\, mA$ range,the resistance to be connected as a shunt is .............. $\Omega$.

$A$ galvanometer is used in a laboratory for detecting the null point in electrical experiments. If, on passing a current of $6 \ mA$, it produces a deflection of $2^{\circ}$, its figure of merit is close to (each division is $1^{\circ}$):

In the adjoining circuit diagram,the readings of the ammeter and voltmeter are $2 \, A$ and $120 \, V$,respectively. If the value of $R$ is $75 \, \Omega$,then the voltmeter resistance will be (in $\Omega$):

The deflection in a moving coil galvanometer of resistance $45 \Omega$ falls from $30$ divisions to $3$ divisions. The length of the shunt wire required to convert the galvanometer into an ammeter is [specific resistance of the material of the shunt wire $= 5 \times 10^{-7} \Omega m$ and area of cross-section of the wire $= 4 \times 10^{-7} m^2$]. (in $m$)

The resistance of a galvanometer is $50\, \Omega$ and the maximum current which can be passed through it is $0.002\, A$. What resistance must be connected to it in order to convert it into an ammeter of range $0 - 0.5\, A$?

The scale of a galvanometer is divided into $100$ equal divisions. It has a current sensitivity of $10 \text{ div./mA}$ and a voltage sensitivity of $4 \text{ div./mV}$. The resistance of the galvanometer is: (in $\Omega$)