$A$ galvanometer has a coil of resistance $100 \Omega$ showing a full scale deflection at $50 \mu A$. The resistance that should be added to use it as an ammeter of range $10 \text{ mA}$ is

If the current is doubled,the deflection is also doubled in

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 has $50$ turns and each turn has an area $2 \times 10^{-4} \ m^2$. The magnetic field produced by the magnet inside the galvanometer is $0.02 \ T$. The torsional constant of the suspension wire is $10^{-4} \ N \ m \ rad^{-1}$. When a current flows through the galvanometer,a full-scale deflection occurs if the coil rotates by $0.2 \ rad$. The resistance of the coil of the galvanometer is $50 \ \Omega$. This galvanometer is to be converted into an ammeter capable of measuring current in the range $0-1.0 \ A$. For this purpose,a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance,in ohms,is:

To convert a galvanometer into an ammeter,one needs to connect a

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