10. Given $y=\sqrt{x-1}+\sqrt{4-x}$ (where $x$ and $y$ are real numbers). Then the difference between the maximum and minimum values of $y$ is ( ).
(A) $\sqrt{6}-3$
(B) 3
(C) $\sqrt{5}-\sqrt{3}$
(D) $\sqrt{6}-\sqrt{3}$
10.D.
$$
\begin{array}{l}
y \geqslant 0, 1 \leqslant x \leqslant 4 . \\
y^{2}=3+2 \sqrt{-x^{2}+5 x-4} \\
=3+2 \sqrt{-\left(x-\frac{5}{2}\right)^{2}+\frac{9}{4}},
\end{array}
$$

Then $3 \leqslant y^{2} \leqslant 6$.
Since $y \geqslant 0$, therefore, $\sqrt{3} \leqslant y \leqslant \sqrt{6}$.