Final answer:
To avoid damaging the galvanometer, a minimum series resistance of 2975 ohms is needed when using two 1.5 V batteries connected in series.
Step-by-step explanation:
To calculate the minimum resistance R that must be placed in series with a galvanometer to avoid damaging it, we use Ohm's law (V = IR), where V is the voltage, I is the current, and R is the resistance. In this case, two 1.5 V batteries connected in series provide a total voltage of 3.0 V. Since the galvanometer's full scale deflection is 1 m/s (which likely means 1 mA for current, assuming 'm/s' is a typo and should read 'mA/s'), the maximum permissible current is 0.001 A (1 mA).
We must also take into account the galvanometer's internal resistance. Assuming it is the same as in the given examples, the galvanometer has an internal resistance of 25 ohms. Using Ohm's law, the total resistance for the circuit, when the full-scale current flows through it, is given by R_total = V/I. Substituting the given values, we obtain R_total = 3 V / 0.001 A = 3000 ohms.
The minimum resistance R needed in series with the galvanometer can be found by subtracting the galvanometer's internal resistance from the total resistance: R_min = R_total - R_galvanometer = 3000 ohms - 25 ohms = 2975 ohms. Therefore, the resistor that needs to be placed in series with the galvanometer to avoid damaging it by exceeding the full-scale deflection must have a resistance of at least 2975 ohms.