$A$ galvanometer having internal resistance $10 \Omega$ requires $0.01 \ A$ for a full scale deflection. To convert this galvanometer to a voltmeter of full-scale deflection at $120 \ V$, we need to connect a resistance of

$A$ galvanometer of resistance $200 \Omega$ is to be converted into an ammeter. The value of shunt resistance which allows $3 \%$ of the main current through the galvanometer is equal to (nearly) (in $\Omega$)

In an ammeter,$0.25 \%$ of the main current passes through the galvanometer. If the resistance of the galvanometer is $G$,the resistance of the ammeter will be:

$A$ moving coil galvanometer has a resistance of $50\,\Omega$ and gives full-scale deflection for $10\,mA$. How could it be converted into an ammeter with a full-scale deflection for $1\,A$?

The circular coil of a galvanometer has $50$ turns. The coil has a radius of $1 \ cm$. The coil is placed in a radial magnetic field of $0.010 \ T$. The torsion coefficient of the spring on which the coil hangs is $3 \times 10^{-7} \ N-m-rad^{-1}$. Determine the deflection of the galvanometer's pointer if the current through the coil is $1.0 \ mA$. (in $^{\circ}$)

When a current of $5 \ mA$ is passed through a galvanometer having a coil of resistance $15 \ \Omega$,it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range $0 - 10 \ V$ is