A voltmeter has a range $0-V$ with a series resistance R. With a series resistance 2R, the range is $0-V\text{'}.$ The correct relation between V and V' is

For conversion of galvanometer (of resistances) into voltmeter, a resistance is connected in series.

$\begin{array}{l}\therefore {\mathrm{i}}_{\mathrm{g}}=\frac{\mathrm{V}}{\mathrm{R}+\mathrm{G}}\mathrm{and}{\mathrm{i}}_{\mathrm{g}}=\frac{{\mathrm{V}}^{\text{'}}}{2\mathrm{R}+\mathrm{G}}\\ \Rightarrow \frac{\mathrm{V}}{\mathrm{R}+\mathrm{G}}=\frac{{\mathrm{V}}^{\text{'}}}{2\mathrm{R}+\mathrm{G}}\\ \Rightarrow \frac{{\mathrm{V}}^{\text{'}}}{\mathrm{V}}=\frac{2\mathrm{R}+\mathrm{G}}{\mathrm{R}+\mathrm{G}}=\frac{2\left(\mathrm{R}+\mathrm{G}\right)-\mathrm{G}}{\left(\mathrm{R}+\mathrm{G}\right)}\\ =2-\frac{\mathrm{G}}{\left(\mathrm{R}+\mathrm{G}\right)}\Rightarrow {\mathrm{V}}^{\text{'}}=2\mathrm{V}-\frac{\mathrm{VG}}{\left(\mathrm{R}+\mathrm{G}\right)}\Rightarrow {\mathrm{V}}^{\text{'}}<2\mathrm{V}\end{array}$