First, we calculate that there are a total of $4\cdot4=16$ possibilities. Now, we list all of two-digit perfect squares. $64$ and $81$ are the only ones that can be made using the spinner. Consequently, there is a $\frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}}$ probability that the number formed by the two spinners is a perfect square.
~MathFun1000
The arrows on the two spinners shown below are spun. Let the number $N$  equal $10$ times the number on Spinner $\text{A}$, added to the number on Spinner $\text{B}$. What is the probability that $N$ is a perfect square number?

$\textbf{(A)} ~\dfrac{1}{16}\qquad\textbf{(B)} ~\dfrac{1}{8}\qquad\textbf{(C)} ~\dfrac{1}{4}\qquad\textbf{(D)} ~\dfrac{3}{8}\qquad\textbf{(E)} ~\dfrac{1}{2}$