Final answer:
To convert a 25.0-Ω galvanometer with 50.0-µA sensitivity to a voltmeter for a 0.100-V full-scale reading, an additional resistance of 1975 Ω is required. This value is derived from Ohm's law. The closest given option is 10.0 kΩ, which appears to be a misprint or misunderstanding.
Step-by-step explanation:
The question involves finding the additional resistance needed to convert a galvanometer into a voltmeter. Given a 25.0-Ω galvanometer with 50.0-µA sensitivity, and requiring a 0.100-V full-scale reading, we can calculate the necessary series resistance using Ohm's law. The full-scale deflection of the galvanometer occurs at its maximum sensitivity current, which is 50.0 µA. Using the formula V = IR, where V is the voltage, I is the current, and R is the resistance, we can calculate the total resistance required for the voltmeter.
To achieve a 0.100-V reading at 50.0 µA, the total resistance must be R_total = V / I = 0.100 V / 50.0 µA = 0.100 V / 0.000050 A = 2000 Ω. Since the galvanometer already has a resistance of 25.0 Ω, the additional series resistance needed is R_series = R_total - R_galvanometer = 2000 Ω - 25.0 Ω = 1975 Ω or 1.975 kΩ.
However, among the given options, the closest value is (b) 10.0 kΩ. Since the given options do not include 1.975 kΩ, and the question might have a typo or miscalculation, you would normally clarify this with the student or instructor. Nevertheless, assuming the options are correct, you would logically select the option that's closest to the required value.