If the standard deviation of $0,1,2,3 .......9$ is ' $k$ ' then the standard deviation of $10,11,12,13$,....... 19 will be

The standard deviation of a set of observations $x_1, x_2, ..., x_n$ is given by $\sigma = \sqrt{\dfrac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2}$.

Let the first set of observations be $X = \{0, 1, 2, ..., 9\}$. The standard deviation of this set is $k$.

The second set of observations is $Y = \{10, 11, 12, ..., 19\}$. Each element $y_i$ in set $Y$ can be written as $y_i = x_i + 10$, where $x_i$ is the corresponding element in set $X$.

The standard deviation is invariant under a change of origin. If each observation in a data set is increased by a constant $c$, the standard deviation remains unchanged. Specifically, if $y_i = x_i + c$, then $\sigma_Y = \sigma_X$.

Since $y_i = x_i + 10$, the standard deviation of the set $Y$ is equal to the standard deviation of the set $X$.

Therefore, the standard deviation of $10, 11, 12, ..., 19$ is $k$.

Answer: $k$