E is definitely wrong. If the volume is already associated with a running container, the second command should just attach to that container, not fail.
D sounds plausible, but I'm not sure why the original container image data would be available in both containers. Hmm, I'll have to think this through a bit more.
I think the correct answer is C. Both containers share the contents of the data volume, have full permissions to alter its content and mutually see their respective changes.
Chan
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