A galvanometer has resistance G ohms and range v volt. The value of resistance used in series to convert into a voltmeter of range nV is:

Hint: A galvanometer is an electronic device which is used for the detection and indication of electric current. Whereas, voltmeter is a device which is used for measuring the electric potential difference between two points in a circuit. Here we have to find the value of the resistance in series that is responsible for converting the galvanometer into voltmeter.

Complete step by step answer:
Find the resistance:
We know the relation between voltage, current and resistor:
$V = IR$;
Write the above formula w.r.t the current I in regard with the galvanometer:
\[{I_g} = \dfrac{V}{G}\]; …(here G = Resistance of Galvanometer)
Now, according to the question a resistance is added in series with the galvanometer to convert it into voltmeter of range nV. So, the total resistance would become.
\[V = ({R_s} + G){I_g}\]; …(here\[{R_s}\]= Resistance in series)
Put the value of \[{I_g}\]in the above relation and solve:
\[V = ({R_s} + G)\dfrac{V}{G}\];
Here the V will become nV:
\[\Rightarrow nV = ({R_s} + G)\dfrac{V}{G}\];
Cancel out the common factor:
\[ \Rightarrow n = ({R_s} + G)\dfrac{1}{G}\];
\[ \Rightarrow nG = ({R_s} + G)\];
Take the G on the RHS to LHS:
 \[ \Rightarrow nG - + G = {R_s}\];
The resistance in series would be:
\[ \Rightarrow {R_s} = G\left( {n - 1} \right)\]

The resistance used in series to convert into a voltmeter of range nV is \[{R_s} = G\left( {n - 1} \right)\].

Note:
Here apply ohm’s law and put in place of resistance and current the value of galvanometric current and resistance and since there is an additional current which is in series with the galvanometric resistance the two resistances will add together and get converted into the voltmeter.