The sensitivity of a galvanometer that measures current is decreased by $\frac{1}{40}$ times by using a shunt resistance of $10 \Omega$. Then,the value of the resistance of the galvanometer is (in $Omega$)

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

$A$ 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 has a deflection of $25$ divisions for a current of $0.1 \ A$ flowing through it. Then the current sensitivity of the galvanometer is (in $div/A$)

$A$ 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:

$A$ galvanometer with $50$ divisions on the scale has a resistance of $25 \Omega$. $A$ current of $2 \times 10^{-4} \text{ A}$ gives a deflection of one scale division. The additional series resistance required to convert it into a voltmeter reading up to $25 \text{ V}$ is $.... \Omega$.