The galvanometer in task-1 will detect an imbalance in the resistance when the current is applied, however the new galvanometer with an internal resistance of 500 Ω will not be able to detect the sa

Task-1: Given: Voltage = 5 V Internal Resistance of Battery (resistance of circuit) = negligible Resistance of Galvanometer = 100 Ω Sensitivity of Galvanometer = 10 mm/µA For a deflection to be visible in the Galvanometer, a current has to pass through it. This current can be calculated using Ohm's law: I=V/R Given that the resistance under measurement has an unbalance of 5 Ω and we're considering the resistance of the galvanometer, the total resistance in the circuit will be 105 Ω. So, Current, I = 5V / 105Ω=0.0476 A =47.6 mA Now, given that the Galvanometer has a sensitivity of 10mm/µA, the deflection on the Galvanometer, for the above calculated current, will be : Deflection = 10mm/µA * 47.6 µA = 476 mm Task-2: In this case, we have the same conditions as Task 1, but the internal resistance of the Galvanometer is now 500Ω and its sensitivity is 1mm/µA. So, the total resistance in the circuit now will be 505 Ω (500Ω from Galvanometer + 5Ω unbalance in resistance). Hence the current passing through the circuit will be, I = 5V / 505Ω=0.0099 A =9.9 mA And the deflection on the Galvanometer will be Deflection = 1mm/µA * 9.9 µA =9.9 mm As mentioned, if a deflection of 1mm can be observed in the Galvanometer, therefore a deflection of about 9.9 mm indicates that this new Galvanometer is capable of detecting 5Ω unbalance in the resistance under measurement.