jmc

First of all, we notice that $11=1+10,$ and so we write $11^n$ as follows: $$(1+10{)}^n=(\genfrac{}{}{0ex}{}{n}{0})\cdot 1^n+(\genfrac{}{}{0ex}{}{n}{1})\cdot 1^{n-1}\cdot 10^1+(\genfrac{}{}{0ex}{}{n}{2})\cdot 1^{n-2}\cdot 10^2+\cdots$$ We can see that every term after the first two in our expansion has at least two powers of $10,$ therefore they will not contribute to the tens digit of anything. Meanwhile, the first term is always $1,$ and the second term can be simplified to $10n.$

Therefore, we have: $$\begin{array}{rl}& 11^1+11^2+11^3+\cdots +11^9\\ & \equiv (1+10)+(1+20)+\cdots +(1+90)(\mathrm{mod}100).\\ & \equiv 459\equiv 59(\mathrm{mod}100).\end{array}$$ Thus, the tens digit must be $\boxed{5}.$