11. In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, the measure of the dihedral angle $A-B D_{1}-A_{1}$ is
11. $60^{\circ}$ Connect $D_{1} C$, draw $C E \perp B D_{1}$, with the foot of the perpendicular at $E$, extend $C E$ to intersect $A_{1} B$ at $F$, then $F E \perp B D_{1}$, connect $A E$,
By the symmetry of the cube, $A E \perp B D_{1}$, and $A E=C E$,
Therefore, $\angle F E A$ is the plane angle of the dihedral angle $A-B D_{1}-A_{1}$.
Connect $A C$, let $A B=1$, then $A C=A D_{1}=\sqrt{2}, B D_{1}=\sqrt{3}$.
In the right triangle $\triangle A B D_{1}$, $A E=\frac{A B \cdot A D_{1}}{B D_{1}}=\frac{\sqrt{2}}{\sqrt{3}}$,
In $\triangle A E C$, $\cos \angle A E C=\frac{A E^{2}+C E^{2}-A C^{2}}{2 A E \cdot C E}=\frac{2 A E^{2}-A C^{2}}{2 A E^{2}}$
$$
=\frac{(4 / 3)-2}{4 / 3}=-\frac{1}{2} .
$$

Thus, $\angle A E C=120^{\circ}$, and $\angle F E A$ is the supplementary angle of $\angle A E C$, so $\angle F E A=60^{\circ}$ is the answer.