$A$ 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 moving coil galvanometer $(MCG)$ is converted into a voltmeter,the series resistance is '$n$' times the resistance of the galvanometer. How many times the original voltage range of the $MCG$ is the voltmeter now capable of measuring?

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.

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 $\Omega$)

In an ammeter,$0.25 \%$ of the main current passes through the galvanometer. If the resistance of the galvanometer is $G$,the resistance of the ammeter will be:

$A$ voltmeter has a resistance of $2000 \, \Omega$ and it can measure up to $2 \, V$. If we want to increase its range to $10 \, V$, then the required resistance in series will be ........... $\Omega$.