If $\overrightarrow{a} .\overrightarrow{ b} = 0$ and $\overrightarrow{a}+\overrightarrow{b}$ makes an angle 60° with $\overrightarrow{a}$, then

$$\begin{gathered} {\left|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\right|}^2={\left|\overrightarrow{\mathrm{a}}\right|}^2+{\left|\overrightarrow{\mathrm{b}}\right|}^2+2\overrightarrow{\mathrm{a}}.\overrightarrow{b} \\ \Rightarrow {\left|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\right|}^2={\left|\overrightarrow{\mathrm{a}}\right|}^2+{\left|\overrightarrow{\mathrm{b}}\right|}^2 \\ and \\ (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\cdot \overrightarrow{\mathrm{a}}=|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\parallel \overrightarrow{\mathrm{a}}|\cos {60}^{\circ } \\ \Rightarrow \stackrel{ \to }{\mathrm{a}}.\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}}.\overrightarrow{b}=|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\parallel \overrightarrow{\mathrm{a}}|\left(\frac{1}{2}\right) \\ \Rightarrow {\left|\overrightarrow{\mathrm{a}}\right|}^2+0=|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\parallel \overrightarrow{\mathrm{a}}|\left(\frac{1}{2}\right) \\ \Rightarrow 2\left|\overrightarrow{\mathrm{a}}\right|=\left|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\right| \\ \Rightarrow 4|\overrightarrow{\mathrm{a}}{|}^2={\left|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\right|}^2 \\ \Rightarrow 4|\overrightarrow{\mathrm{a}}{|}^2=|\overrightarrow{\mathrm{a}}{|}^2+|\overrightarrow{\mathrm{b}}{|}^2 \\ \Rightarrow \left|\overrightarrow{\mathrm{b}}\right|=\sqrt{3}\left|\overrightarrow{\mathrm{a}}\right| \end{gathered}$$