Why is the total resistance of a voltmeter kept very high?

In the given figure,an ammeter $A$ consists of a $240 \Omega$ coil connected in parallel to a $10 \Omega$ shunt. The reading of the ammeter is . . . . . . $mA$.

Derive the expression for the shunt resistance required to increase the current-measuring capacity of a galvanometer by a factor of $n$.

$A$ galvanometer of resistance $G$ can be converted into a voltmeter of range $V$ by connecting a resistance $R$ in series with it. The series resistance required to change its range to $\frac{V}{3}$ is

$A$ galvanometer has resistance $G \ \Omega$ and $I_g$ is the current flowing through it which produces full-scale deflection. $S_1$ is the value of the shunt which converts it into an ammeter of range $0$ to $3I$,and $S_2$ is the shunt value which converts it into an ammeter of range $0$ to $4I$. The ratio $S_2:S_1$ is:

$A$ galvanometer of resistance $100\,\Omega$ has $50$ divisions on its scale and a sensitivity of $20\,\mu A/\text{division}$. It is to be converted into a voltmeter with three ranges: $0-2\,V$,$0-10\,V$,and $0-20\,V$. The appropriate circuit to do so is: