2. Given a rhombus $A B C D$ with side $A B=10$, diagonal $B D=12$, and 100 different points $P_{1}, P_{2}$, $\cdots, P_{100}$ on $B D$. Draw $P_{i} E_{i} \perp A B$ at $E_{i}$, and $P_{i} F_{i} \perp A D$ at $F_{i}$. Then
$$
P_{1} E_{1}+P_{1} F_{1}+P_{2} E_{2}+P_{2} F_{2}+\cdots+P_{100} E_{100}+P_{100} F_{100}
$$
is $(\quad)$.
(A) 9.6
(B) 1920
(C) 960
(D) cannot be determined
2. C.

As shown in Figure 5, extend $E_{i} P_{i}$ to intersect $C D$ at point $G_{i}$. Then,
$$
P_{i} G_{i} \perp C D \text {. }
$$

Thus, $P_{i} F_{i}=P_{i} G_{i}$. Connect $A C$ to intersect $B D$ at point $O$. Therefore,
$$
\begin{array}{l}
A C \perp B D, \\
B O=\frac{1}{2} B D=6 .
\end{array}
$$

Then $A O=\sqrt{A B^{2}-B O^{2}}=\sqrt{10^{2}-6^{2}}=8$,
$$
A C=2 A O=16 \text {. }
$$

From $S_{\text {parallelogram }}=A B \cdot E_{i} G_{i}=\frac{1}{2} A C \cdot B D$, we get
$$
10 E_{i} G_{i}=\frac{1}{2} \times 16 \times 12 \text {. }
$$

Solving for $E_{i} G_{i}=9.6$, thus
$$
\begin{array}{l}
P_{i} E_{i}+P_{i} F_{i}=P_{i} E_{i}+P_{i} G_{i} \\
=9.6(i=1,2, \cdots, 100) .
\end{array}
$$

Therefore, $P_{1} E_{1}+P_{1} F_{1}+P_{2} E_{2}+P_{2} F_{2}+\cdots+$
$$
P_{100} E_{100}+P_{100} F_{100}=9.6 \times 100=960 .
$$