jmc

If $101\le x\le 110$, then notice that $10=\sqrt{100}<\sqrt{x}<11=\sqrt{121}$. Thus, $\lfloor \sqrt{x}{\rfloor }^2=10^2=100$. The desired sum is then $(101-100)+(102-100)+\cdots +(110-100)=1+2+\cdots +10=\frac{10\cdot 11}{2}=\boxed{55}$.