$A$ galvanometer of resistance $G$ is shunted by a resistance $S$. To keep the main current in the circuit unchanged,the resistance to be put in series with the galvanometer is

Only $4 \%$ of the total current in the circuit passes through a galvanometer. If the resistance of the galvanometer is $G$,then the shunt resistance connected to the galvanometer is:

In an experiment to determine the figure of merit of a galvanometer by the half-deflection method,a student constructed the following circuit. He unplugged a resistance of $5200 \ \Omega$ in $R$. When $K_1$ is closed and $K_2$ is open,the deflection observed in the galvanometer is $26 \ \text{div}$. When $K_2$ is also closed and a resistance of $90 \ \Omega$ is removed in $S$,the deflection becomes $13 \ \text{div}$. The resistance of the galvanometer is nearly: (in $\Omega$)

$A$ galvanometer has $30$ divisions and a current sensitivity of $20 \, \mu A$ per division. Its resistance is $25 \, \Omega$. How will you convert this galvanometer into a voltmeter of $1 \, V$ range? (Find the series resistance in $\Omega$)

$A$ 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 $= 5 \times 10^{-7} \Omega\text{-m}$)

$A$ student is provided with a variable voltage source $V$,a test resistor $R_T=10\,\Omega$,two identical galvanometers $G_1$ and $G_2$,and two additional resistors,$R_1=10\,M\,\Omega$ and $R_2=0.001\,\Omega$. For conducting an experiment to verify Ohm's law,the most suitable circuit is: