Find the resistance of the frustum of a cone made with a material of uniform thickness and resistivity 'ro'. lets say 'Z'. The radii of the smaller and larger flat bases are 'L' and '2L' respectively. Height = 'L'. The two flat bases are covered with a zero resistance material. The resistance is measured between two terminals, one lead at the centre of the smaller flat base and the other lead at the centre of the larger flat base.
They must be joking. . .
You do have to know the formula for the volume of a conic frustrum or be able to calculate it on the spot.
Z = ro*L/A, where A is the cross-sectional area of a volume of length L and ro is the volume or bulk resistivity, ρ [rho], of the material. Z is resistive, not reactive, since ρ is resistive.
The volume of this conic frustrum, V, equals (PI*L/3)*(4*[L^2] + 2*[L^2] + [L^2])
Assume this conic frustrum is a cylinder of length L and average cross-sectional area A, where A*L = V. So, A = V/L.
Therefore, I believe that Z = ro*[L^2]/V.
Once you have Z, if V is a DC voltage source and C is the value in farads of a capacitor and Z is the value in ohms of the R from above the voltage across the capacitor Vc = V*{1-e^(-t/[Z*C])}; again, Z is resistive.
These are probably questions from a PE exam for electrical engineers.