If an ammeter is to be used in place of a galvanometer,then we must connect:

Given below are two statements: One is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A:$ For measuring the potential difference across a resistance of $600\,\Omega$,the voltmeter with resistance $1000\,\Omega$ will be preferred over a voltmeter with resistance $4000\,\Omega$.
Reason $R:$ $A$ voltmeter with higher resistance will draw smaller current than a voltmeter with lower resistance.
In the light of the above statements,choose the most appropriate answer from the options given below.

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]

The resistance of a $1\, A$ ammeter is $0.018\,\Omega$. To convert it into a $10\, A$ ammeter,the shunt resistance required 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 $m$)

$A$ galvanometer has a resistance of $25 \, \Omega$ and a maximum of $0.01 \, A$ current can be passed through it. In order to change it into an ammeter of range $10 \, A$, the shunt resistance required is