Two springs of force constants  100Nm^-1 and  200Nm^-1 are stretched by the same force. Their PE will be in the ratio

$\begin{array}{l}{\mathrm{F}}_1={\mathrm{k}}_1{\mathrm{x}}_1,{\mathrm{F}}_2={\mathrm{k}}_2{\mathrm{x}}_2\\ \text{ But }{\mathrm{F}}_1={\mathrm{F}}_2\\ \therefore {\mathrm{k}}_1{\mathrm{x}}_1={\mathrm{k}}_2{\mathrm{x}}_2\Rightarrow 100{\mathrm{x}}_1=200{\mathrm{x}}_2\\ {\mathrm{x}}_1=2{\mathrm{x}}_2\\ {\mathrm{PE}}_1=\frac{1}{2}{\mathrm{k}}_1{\mathrm{x}}_1^2={\mathrm{PE}}_2=\frac{1}{2}{\mathrm{k}}_2{\mathrm{x}}_2^2\\ \frac{{\mathrm{PE}}_1}{{\mathrm{PE}}_2}=\frac{{\mathrm{k}}_1{\mathrm{x}}_1^2}{{\mathrm{k}}_2{\mathrm{x}}_2^2}=\frac{100\times {\mathrm{x}}_1^2}{200\times \frac{{\mathrm{x}}_1^2}{4}}=\frac{2}{1}=2:1\end{array}$