The equivalent of the Newton's second law for rotational motions is:

where

is the net torque acting on the object

is its moment of inertia

is the angular acceleration of the object.
Re-arranging the formula, we get

and since we know the net torque acting on the (vase+potter's wheel) system,

, and its angular acceleration,

, we can calculate the moment of inertia of the system:
Your brother would be described as a speeder.
This is a tough one.
In A, B, and C, there are various routes to get from one end to the other end through 2 or 3 capacitors.
D is the only configuration where it's possible to get from one end to the other through only 1 of them.
I'm not totally confident, but I think D is the one that's not like the others.