AP Physics 1 Exam Question:

A student is investigating the resistance of a wire. The wire is 2 meters long and has a diameter of 0.5 millimeters. The wire is made of copper, which has a resistivity of 1.7 x 10^-8 Ω*m. The student connects the wire to a battery with a voltage of 9 volts and measures a current of 0.5 amperes flowing through it.

a) Calculate the resistance of the wire in ohms.

b) If the length of the wire is doubled while keeping the diameter constant, how does the resistance change? Explain why.

c) If the diameter of the wire is doubled while keeping the length constant, how does the resistance change? Explain why.

Answer:

a) The resistance of a wire is given by the formula:

R = (ρ * L) / A

where R is the resistance, ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.

Given: Length of the wire, L = 2 meters Diameter of the wire, d = 0.5 millimeters = 0.0005 meters Resistivity of copper, ρ = 1.7 x 10^-8 Ω*m

To find the cross-sectional area of the wire, we need to first find the radius:

r = d/2 = 0.0005/2 = 0.00025 meters

A = π * r^2 = π * (0.00025)^2 = 1.963 x 10^-7 m^2

Now we can substitute the values into the resistance formula:

R = (ρ * L) / A = (1.7 x 10^-8 * 2) / (1.963 x 10^-7) = 0.173 Ω

Therefore, the resistance of the wire is 0.173 ohms.

b) If the length of the wire is doubled while keeping the diameter constant, the resistance will also double. This is because the resistance is directly proportional to the length of the wire (assuming the resistivity and cross-sectional area remain constant). As the length increases, there is more wire for the electrons to flow through, resulting in an increased resistance.

c) If the diameter of the wire is doubled while keeping the length constant, the resistance will decrease by a factor of 4. This is because the resistance is inversely proportional to the cross-sectional area of the wire (assuming the resistivity and length remain constant). As the diameter doubles, the cross-sectional area increases by a factor of 4 (since it is proportional to the square of the radius), resulting in a decreased resistance.