If the variance of 10 natural numbers 1, 1, 1, , 1, k is less than...

${\sigma}^{2} = \dfrac{Σ{x}^{2}}{n} - {\left(\dfrac{Σx}{n}\right)}^{2}$

$= \dfrac{9 + {k}^{2}}{10} - {\left(\dfrac{9 + k}{10}\right)}^{2} < 10$

$90 + 10{k}^{2} - 81 - {k}^{2} - 18k < 1000$

$9{k}^{2} - 18k - 991 < 0$

${k}^{2} - 2k < \dfrac{991}{9}$

${\left(k - 1\right)}^{2} < \dfrac{1000}{9}$

$\dfrac{- 10\sqrt{10}}{3} < k - 1 < \dfrac{10\sqrt{10}}{3}$

$k < \dfrac{10\sqrt{10}}{3} + 1$

$k \le 11$

Maximum value of $k$ is $11 .$