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 $E = 2 \text{ V}$]

Explain the solution to the difficulties that arise when using a galvanometer directly as an ammeter.

$A$ galvanometer $G$ deflects full scale when a potential difference of $0.50 \ V$ is applied. The internal resistance of the galvanometer $r_g$ is $25 \ \Omega$. An ammeter is constructed by incorporating the galvanometer and an additional shunt resistance $R_S$. The ammeter deflects full scale when a measurement of $2.0 \ A$ is made. The resistance $R_S$ is closest to : ................. $\Omega$

$A$ galvanometer,having a resistance of $50 \Omega$,gives a full scale deflection for a current of $0.05 \text{ A}$. The length in metre of a resistance wire of area of cross-section $2.97 \times 10^{-2} \text{ cm}^2$ that can be used to convert the galvanometer into an ammeter which can read a maximum of $5 \text{ A}$ current is: (Specific resistance of the wire $= 5 \times 10^{-7} \Omega\text{-m}$)

Why should the resistance of an ammeter be as low as possible?

State the principle of a moving coil galvanometer.