Resistance is given by $$R = \rho L/A$$, where $$A = \pi r^2$$. For Wire A, $$R_A = \rho L/(\pi r^2)$$. For Wire B, $$L_B=2L$$ and $$r_B=r/2$$, so $$A_B = \pi(r/2)^2 = \pi r^2/4$$. Thus, $$R_B = \rho(2L)/(\pi r^2/4) = 8(\rho L/(\pi r^2)) = 8R_A$$. The ratio $$R_B/R_A$$ is 8.

From Ohm's Law, $$V=IR$$. A graph of $$V$$ (y-axis) versus $$I$$ (x-axis) for an ohmic resistor is a straight line with slope $$m = \Delta V / \Delta I$$. This slope is equal to the resistance $$R$$. Therefore, the resistance of the component is $$200,\Omega$$.

The volume of the wire is $$V = A \cdot L$$. If the volume is constant, then $$A \propto 1/L$$. Resistance is given by $$R = \rho L/A$$. Since $$A \propto 1/L$$, we can say $$R \propto L/(1/L) = L^2$$. If the length $$L$$ is doubled, the resistance will increase by a factor of $$2^2 = 4$$.

A non-ohmic material is one that does not follow Ohm's Law, which states that the ratio of voltage to current (the resistance) is constant. Therefore, for a non-ohmic material, the resistance changes depending on the operating conditions like voltage or current.

To determine if the element is ohmic, we calculate the ratio $$R = V/I$$ for each data point: $$2.0/0.50 = 4.0,\Omega$$, $$4.0/1.00 = 4.0,\Omega$$, $$6.0/1.50 = 4.0,\Omega$$, and $$8.0/2.00 = 4.0,\Omega$$. Since the ratio is constant, the element is ohmic with a resistance of $$4.0,\Omega$$.

For most metallic conductors, resistivity increases with temperature. This is described by a positive temperature coefficient of resistivity. Since resistance $$R$$ is directly proportional to resistivity $$\rho$$ ($$R = \rho L/A$$), an increase in temperature will cause an increase in the resistance of the metallic resistor.

The resistance $$dR$$ of an infinitesimal slice of length $$dx$$ is $$dR = \rho(x) \frac{dx}{A}$$. To find the total resistance, we integrate from $$x=0$$ to $$x=L$$: $$R = \int_0^L \frac{\rho_0}{A}(1 + \frac{x^2}{L^2}) dx = \frac{\rho_0}{A} \left[x + \frac{x^3}{3L^2}\right]_0^L = \frac{\rho_0}{A} \left(L + \frac{L^3}{3L^2}\right) = \frac{\rho_0}{A} \left(L + \frac{L}{3}\right) = \frac{4\rho_0 L}{3A}$$.

First, calculate the resistance using Ohm's Law: $$R = V/I = 1.0,\text{V} / 4.0,\text{A} = 0.25,\Omega$$. Next, calculate the cross-sectional area: $$A = \pi r^2 = \pi(0.50 \times 10^{-3},\text{m})^2 \approx 7.85 \times 10^{-7},\text{m}^2$$. Finally, use the resistance formula to find resistivity: $$\rho = RA/L = (0.25,\Omega)(7.85 \times 10^{-7},\text{m}^2) / (2.0,\text{m}) \approx 9.8 \times 10^{-8},\Omega \cdot \text{m}$$.

This question tests understanding of resistance, resistivity, and Ohm's Law in AP Physics C: Electricity and Magnetism. Ohm's Law V = IR shows that for constant voltage, current and resistance have an inverse relationship, expressed as I = V/R. In the lightbulb circuit example, when the filament's resistance increases while the applied voltage remains constant, the current through the bulb must decrease to satisfy Ohm's Law. Choice B is correct because it accurately states that current decreases as resistance increases when voltage is constant, following directly from the mathematical relationship I = V/R. Choice A is incorrect as it suggests a direct relationship between current and resistance, which would violate Ohm's Law and conservation principles. Students should understand that increased resistance means more opposition to current flow, like a dimmer switch increasing resistance to reduce bulb brightness. Practice with real circuits helps: a 60W bulb has lower resistance than a 40W bulb, so it draws more current at the same voltage.

This question tests understanding of resistance, resistivity, and Ohm's Law in AP Physics C: Electricity and Magnetism. Ohm's Law V = IR rearranges to I = V/R, establishing that current is inversely proportional to resistance when voltage remains constant in a circuit. In the lightbulb example, if the filament's resistance increases while the household circuit maintains constant voltage, the current through the bulb must decrease according to the relationship I = V/R. Choice B is correct because it states that current decreases and provides the correct mathematical relationship I = V/R that governs this inverse proportionality. Choice D is incorrect because it misinterprets Ohm's Law - current does not become equal to resistance; rather, current equals voltage divided by resistance, and they have different units (amperes vs ohms). Students should practice dimensional analysis with Ohm's Law to verify relationships and avoid unit confusion. Emphasize that in household circuits, voltage is typically fixed by the power company, making resistance the primary variable for controlling current in devices.