4. Given that $A B C D-A_{1} B_{1} C_{1} D_{1}$ is a cube with edge length 1, $O_{1}$ is the center of the base $A_{1} B_{1} C_{1} D_{1}$, and $M$ is a point on the edge $B B_{1}$ such that $S_{\triangle D B M}: S_{\triangle O_{1} B_{1} M}=$ 2:3. Then the volume of the tetrahedron $O_{1} A D M$ is ( ).
(A) $\frac{7}{24}$
(B) $\frac{3}{16}$
(C) $\frac{7}{48}$
(D) $\frac{11}{48}$
4.C.

As shown in Figure 1, it is easy to see that $A C \perp$ plane $D_{1} B_{1} B D$. Let $O$ be the center of the base $A B C D$, then $A O \perp$ plane $D O_{1} M$.
$$
\begin{aligned}
& \text { Since } \frac{S_{\triangle D B M}}{S_{\triangle O_{1} B_{1} M}} \\
= & \frac{B D \cdot B M}{O_{1} B_{1} \cdot B_{1} M}=2 \times \frac{B M}{B_{1} M}=\frac{2}{3},
\end{aligned}
$$

Therefore, $\frac{B M}{B_{1} M}=\frac{1}{3}$.
Thus, $B M=\frac{1}{4}, B_{1} M=\frac{3}{4}$.
Hence, $S_{\triangle D O_{1} M}$
$$
\begin{array}{l}
=S_{\text {quadrilateral } D_{1} B_{1} B D}-S_{\triangle D D_{1} O_{1}}-S_{\triangle O_{1} B_{1} M}-S_{\triangle D B M} \\
=\frac{7 \sqrt{2}}{16} .
\end{array}
$$

Then $V_{A-O_{1} M D}=\frac{1}{3} S_{\triangle D O_{1} M} \cdot A O=\frac{7}{48}$.