An ammeter gives full scale deflection when a current of $1.0 \ A$ is passed through it. To convert it into a $10 \ A$ range ammeter,the ratio of its resistance $(G)$ to the shunt resistance $(S)$ will be:

The deflection in a moving coil galvanometer of resistance $45 \Omega$ falls from $30$ divisions to $3$ divisions. The length of the shunt wire required to convert the galvanometer into an ammeter is [specific resistance of the material of the shunt wire $= 5 \times 10^{-7} \Omega m$ and area of cross-section of the wire $= 4 \times 10^{-7} m^2$]. (in $m$)

In order to increase the sensitivity of a moving coil galvanometer,one should decrease

To find the resistance of a galvanometer by the half-deflection method,the following circuit is used with resistances $R_1 = 9970\,\Omega$,$R_2 = 30\,\Omega$,and $R_3 = 0\,\Omega$. The deflection in the galvanometer is $d$. With $R_3 = 107\,\Omega$,the deflection changes to $\frac{d}{2}$. The galvanometer resistance is approximately ............... $\Omega$.

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

In the conversion of a moving coil galvanometer into an ammeter of a required range,what is the resistance of the ammeter so formed? [$S$ = shunt resistance and $G$ = resistance of the galvanometer]