An $O(n^2)$ is faster than an $O(n\log n)$ algorithm for small $n$
Your statement is meaningless, since $O(n^2)$ and $O(n\log n)$ are just upper bounds on the time complexity. If you know that $A \leq 100$ and $B \leq 10$, you have absolutely no idea which is larger, $A$ or $B$.
Let us correct your statement to

If $n&lt;100$ then $\Theta(n^2)$ is more efficient, but if $n \geq 100$ then $\Theta(n\log n)$ is more efficient.

This statement still has no truth value, since we don't know which functions are represented by $\Theta(n^2)$ and by $\Theta(n\log n)$. What we do know is that for large enough $n$, the $\Theta(n\log n)$ algorithm would be more efficient.
Suppose you instantiate your statement with specific functions $f(n) = \Theta(n^2)$ and $g(n) = \Theta(n\log n)$. There is no reason to expect that there is a single crossover point. For example, take the following functions:
$$
f(n) =
\begin{cases}
100 &amp; \text{if } n &lt; 20, \\
n^2 &amp; \text{if } n \ge 20
\end{cases} \\
g(n) =
\begin{cases}
50 &amp; \text{if } n &lt; 10, \\
200 &amp; \text{if } 10 \le n &lt; 20, \\
10^{300} n\log n &amp; \text{if } n \ge 20
\end{cases}
$$
As you can see, $g(n)$ is faster for $n &lt; 10$, and then for (astronomically) large enough $n$.
While such definitions are artificial, in practice we could have (say) $f(n) = 3n^2 + n\log n + 9n + 17$ and $g(n) = 10n\log n + 4\sqrt{n} + 50$, which might have several crossover points (I haven't checked).
Finally, even if there is a single crossover point, there is no reason to expect it to be exactly $n = 100$. The exact crossover point(s) depend on the functions $f(n),g(n)$. The only way to determine it is by running the two algorithms and comparing the running times.