Why does a galvanometer not show full-scale deflection after connecting it in a circuit?

$A$ 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

To find the resistance of a galvanometer by the half-deflection method,the experimental data obtained are given in the table below:

$S. No.$	Resistance $R \ (\Omega)$	Deflection $(\theta)$	Shunt $S \ (\Omega)$	Half deflection $(\theta / 2)$	Galvanometer resistance $(G)$
$1$	$3300$	$30$	$80$	$15$	$G_1$
$2$	$5000$	$20$	$80$	$10$	$G_2$

From the above data,the galvanometer resistance $G$ will be near to: (in $\Omega$)

$A$ moving coil galvanometer of resistance $100 \Omega$ is used as an ammeter using a shunt resistance $0.1 \Omega$. The maximum deflection current in the galvanometer is $100 \mu A$. Find the total current in the circuit so that the ammeter shows maximum deflection. (in $mA$)

The deflection in a moving coil galvanometer falls from $25$ divisions to $5$ divisions when a shunt of $24\ \Omega$ is applied. The resistance of the galvanometer coil will be: (in $\Omega$)

$A$ voltmeter having a resistance of $998 \, \Omega$ is connected to a cell of $e.m.f.$ $2 \, V$ and internal resistance $2 \, \Omega$. The error in the measurement of $e.m.f.$ will be