Final answer:
The filament diameter is reduced approximately by a factor of 0.900. This is derived from the square root of the power reduction factor, which is 0.833, since power is proportional to the square of the diameter.
Step-by-step explanation:
The question is asking us to find the factor by which the diameter of a lightbulb's filament is reduced when the power drawn by the bulb decreases due to evaporative thinning of the filament. Using the power formula for electrical resistance, P = V2 / R, and knowing the resistance of a wire is R = ρL / A, where ρ is the resistivity, L is the length, and A is the cross-sectional area of the wire, we can infer the relationship between power and diameter of the filament (assuming constant resistivity and length).
The power is proportional to 1/R, and since R is proportional to 1/A, P is then proportional to the area of the cross-section of the filament. As the area of a circle A = πd2/4, we see that power is proportional to d2, so when the power is reduced, the diameter must be reduced as well. Since the power goes from 60.0 W to 50.0 W, the reduction factor for power is 50.0/60.0 = 0.833. The diameter reduction factor must be the square root of that, because power scales with the square of the diameter.
Therefore, the diameter reduction factor is the square root of 0.833, which is approximately 0.9129. This is closest to option b. 0.900, representing a 10% reduction in the diameter of the filament.