If the variance of 10 natural numbers 2, 2, 2--------2, k is less than 10, then the maximum possible value of k is

$Mean=\frac{2\left(9\right)+k}{10}=\frac{18+k}{10}$ 
$\sum x_i^2=9{\left(2\right)}^2+k^2=36+k^2$
$\sum _{10}{x}_1^2-{\left(\frac{{\displaystyle \sum x_i}}{10}\right)}^2<10\Rightarrow \frac{36+k^2}{10}-{\left(\frac{18+k}{10}\right)}^2<10$
$360+10k^2<100+324+k^2+36k$
$9k^2-36k-64<0$
$k_{\max }<\frac{36+\sqrt{3600}}{18}$
$<\frac{36+60}{18}=\frac{16}{3}$
k is natural number $\Rightarrow k_{\max }=5$