$A$ 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:

$A$ galvanometer with $30$ divisions has a current sensitivity of $20 \, \mu A/division$. Its resistance is $25 \, \Omega$. How will you convert it into an ammeter to measure $1 \, A$ current?

When a resistance of $100 \Omega$ is connected in series with a galvanometer of resistance $G$,its range is $V$. To double its range,a resistance of $1000 \Omega$ is connected in series. The value of $G$ is (in $\Omega$)

For designing a voltmeter of range $50\,V$ and an ammeter of range $10\,mA$ using a galvanometer which has a coil of resistance $54\,\Omega$ showing a full scale deflection for $1\,mA$ as in figure.
$(A)$ for voltmeter $R \approx 50\,k\Omega$
$(B)$ for ammeter $r \approx 0.2\,\Omega$
$(C)$ for ammeter $r \approx 6\,\Omega$
$(D)$ for voltmeter $R \approx 5\,k \Omega$
$(E)$ for voltmeter $R \approx 500 \Omega$
Choose the correct answer from the options given below

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:

$A$ galvanometer of resistance $100 \Omega$ when connected in series with $400 \Omega$ measures a voltage of up to $10 \ V$. The value of resistance required to convert the galvanometer into an ammeter to read up to $10 \ A$ is $x \times 10^{-2} \Omega$. The value of $x$ is: