$A$ galvanometer of resistance $40 \Omega$ gives a deflection of $10 \text{ divisions per } mA$. There are $50 \text{ divisions}$ on the scale. The maximum current that can pass through the circuit when a shunt resistance of $2 \Omega$ is connected is: (in $\text{ mA}$)

In the circuit (Figure) the current is to be measured. What is the value of the current if the ammeter shown:
$(a)$ is a galvanometer with a resistance $R_{G}=60.00 \; \Omega$;
$(b)$ is a galvanometer described in $(a)$ but converted to an ammeter by a shunt resistance $r_{s}=0.02 \; \Omega$;
$(c)$ is an ideal ammeter with zero resistance?

An ammeter gives full scale deflection when a current of $1.0 \ A$ is passed through it. To convert it into a $10 \ A$ range ammeter,the ratio of its resistance $(G)$ to the shunt resistance $(S)$ will be:

This question has Statement-$I$ and Statement-$II$. Of the four choices given after the statements,choose the one that best describes the two statements.
Statement-$I$: Higher the range,greater is the resistance of an ammeter.
Statement-$II$: To increase the range of an ammeter,an additional shunt needs to be used across it.

$A$ galvanometer has a $50$ division scale. The battery has no internal resistance. It is found that there is a deflection of $40$ divisions when $R = 2400\,\Omega$. The deflection becomes $20$ divisions when the resistance taken from the resistance box is $4900\,\Omega$. Then we can conclude:

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 $..... \mu A / \text{division}$.