How can we convert a galvanometer into an ammeter?

The net resistance of a voltmeter should be large to ensure that

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 $E = 2 \text{ V}$]

$A$ galvanometer coil has a resistance $80 \Omega$ and current for full-scale deflection is $10 \text{ mA}$. $A$ resistance of $920 \Omega$ is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is $0.2 \text{ V}$,the number of divisions on the scale is:

The coil of a galvanometer consists of $100$ turns and an effective area of $1 \, cm^2$. The restoring couple per unit radian is $10^{-8} \, N \cdot m/rad$. The magnetic field between the pole pieces is $5 \, T$. The current sensitivity of this galvanometer will be:

$A$ galvanometer of resistance $20 \Omega$ gives a deflection of $5$ divisions when $1 \text{ mA}$ current flows through it. The galvanometer scale has $50$ divisions. To convert the galvanometer into a voltmeter of range $25 \text{ V}$,we should connect a resistance of