2. For any natural numbers $m, n$ satisfying $m<\sqrt{11} n$, the inequality $11 n^{2} \geqslant m^{2}+k$ always holds. Then the maximum value of $k$ is ( ).
(A) 1
(B) 2
(C) 6
(D) 10
2. B.

Since $m0$.
Also, the remainder of $m^{2}$ divided by 11 can be $0,1,4,9,5$, 3, and $11 n^{2}$ can be divisible by 11, then the minimum value of $11 n^{2}-m^{2}$ is 2.
Since $11 n^{2}-m^{2} \geqslant k$, therefore, the maximum value of $k$ is 2.