The relation between voltage sensitivity $({\sigma _V})$ and current sensitivity $({\sigma _i})$ of a moving coil galvanometer is (Resistance of Galvanometer = $G$)

$A$ voltmeter of resistance $1000\,\Omega$ gives full scale deflection when a current of $100\,mA$ flows through it. The shunt resistance required across it to enable it to be used as an ammeter reading $1\,A$ at full scale deflection is ............... $\Omega$.

$A$ galvanometer of resistance $100\,\Omega$ has $50$ divisions on its scale and a sensitivity of $20\,\mu A/\text{division}$. It is to be converted into a voltmeter with three ranges: $0-2\,V$,$0-10\,V$,and $0-20\,V$. The appropriate circuit to do so 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 $= 5 \times 10^{-7} \Omega\text{-m}$)

$A$ $36 \,\Omega$ galvanometer is shunted by a resistance of $4 \,\Omega$. The percentage of the total current that passes through the galvanometer is (in $\%$)

$A$ voltmeter of resistance $R$ is to be used to measure the potential difference across a resistance $r$. In which case will the reading be nearest to the actual potential difference in the absence of the voltmeter?