For a reaction, $\frac{{\mathrm{K}}_{\mathrm{t}+10}}{{ \mathrm{K}}_{\mathrm{t}}}=\mathrm{x}$. When temperature is increased from $10°\mathrm{C}$ to $100°\mathrm{C}$, rate constant $\mathrm{K})$ increased by a factor of 512 . Then, value of x is

Temperature difference given in the question is:

$$\begin{gathered} \mathrm{\Delta T}={\mathrm{T}}_2-{\mathrm{T}}_1=100-10=90 \\ \frac{{\mathrm{K}}_{{\mathrm{T}}_2}}{{\mathrm{K}}_{{\mathrm{T}}_1}}=512 \\ \frac{{\mathrm{K}}_{\mathrm{t}}+10}{{\mathrm{K}}_{\mathrm{t}}}=\mathrm{x} \end{gathered}$$

Hence, the formula used will be:

$\frac{{\mathrm{K}}_{{\mathrm{T}}_2}}{{\mathrm{K}}_{{\mathrm{T}}_1}}=\mathrm{\mu }\frac{\mathrm{\Delta T}}{10}$

$\mathrm{\mu } =$Temperature coefficient of the reaction

On putting the values, we get:

$$\begin{gathered} \begin{array}{rl} & 512=\mathrm{\mu }\frac{\mathrm{\Delta T}}{10}\\ & 512=\mathrm{\mu }\frac{90}{10}\\ & 2^9={\mathrm{\mu }}^9\end{array} \\ \text{so }\mathrm{\mu }=2 \end{gathered}$$.