The resistance of an ideal voltmeter is . . . . . . .

$A$ galvanometer gives full scale deflection with $0.006 \ A$ current. By connecting it to a $4990 \ \Omega$ resistance,it can be converted into a voltmeter of range $0-30 \ V$. If connected to a $\frac{2n}{249} \ \Omega$ resistance,it becomes an ammeter of range $0-1.5 \ A$. The value of $n$ is:

$A$ 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

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

How are a voltmeter and an ammeter connected in a circuit? Explain.

Which one of the following modifications may increase the sensitivity of a moving coil galvanometer?
$1^{st}$ Way: By using a spring of smaller torsion constant.
$2^{nd}$ Way: By using a smaller coil.
$3^{rd}$ Way: By using a stronger magnet.
$4^{th}$ Way: By using a coil having fewer number of turns.