$A$ galvanometer has resistance '$G$' and range '$V_g$'. How much resistance is required to read voltage up to '$V$' volt?

In a galvanometer,$5 \%$ of the total current in the circuit passes through it. If the resistance of the galvanometer is $G$,the shunt resistance $S$ connected to the galvanometer is

To verify Ohm's law,a student is provided with a test resistor $R_T$,a high resistance $R_1$,a small resistance $R_2$,two identical galvanometers $G_1$ and $G_2$,and a variable voltage source $V$. The correct circuit to carry out the experiment is

$A$ milliammeter of range $10\, mA$ and resistance $9\, \Omega$ is joined in a circuit as shown. The meter gives full-scale deflection for current $I$ when $A$ and $B$ are used as its terminals,i.e.,current enters at $A$ and leaves at $B$ ($C$ is left isolated). The value of $I$ is

$A$ 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:

$A$ galvanometer,having a resistance of $50 \Omega$,gives a full scale deflection for a current of $0.05 \text{ A}$. The length in metre of a resistance wire of area of cross-section $2.97 \times 10^{-2} \text{ cm}^2$ that can be used to convert the galvanometer into an ammeter which can read a maximum of $5 \text{ A}$ current is: (Specific resistance of the wire $= 5 \times 10^{-7} \Omega\text{-m}$)