Final answer:
The ratio of the diameters is 2:1, with the aluminum wire's diameter being twice that of copper to have equal resistance. Calculating current flow through silicon is not possible with information provided.
Step-by-step explanation:
Resistance Ratio and Current through a Silicon Rod
Regarding the ratio of the diameters between aluminum and copper wires with the same resistance per unit length, we must consider the materials' resistivity. The resistance R of a wire is given by R = ρL/A, where ρ is the resistivity, L is the length, and A is the cross-sectional area of the wire. Since the resistance is the same for both wires, and the length is presumably equal, the ratio of their cross-sectional areas, which are directly proportional to the squares of their diameters (A ≤ πd^2/4), will relate inversely to the resistivity. Copper has lower resistivity compared to aluminum, which means the aluminum wire must compensate with a greater diameter to have the same resistance. Accordingly, the answer is b. 2:1, reflecting that the aluminum wire's diameter must be twice that of the copper wire.
For the question about the current flowing through a silicon rod, we utilize Ohm's Law (V = IR), where V is voltage (1.00 × 10^3 V), I is current, and R is the resistance of the rod. Silicon has specific resistivity and requires certain considerations for its conductivity that vary with impurities and temperature. Without details on these variables, the problem cannot be precisely solved. Nonetheless, a pure silicon rod would typically have very high resistance, and thus, the expected current would likely be very low, far less than any of the options like 2.00 A provided in the problem statement. Thus, this part of the problem is left unsolved due to insufficient information.