In a galvanometer, there is a deflection of 10 divisions per mA. The resistance of the galvanometer is 60 Ω. If a shunt of 1 Ω is connected to the galvanometer and there are 100 divisions in all, on the scale of the galvanometer, the maximum current that this galvanometer can read is

610 mA

The correct answer is Option (2) → 610 mA

Given:

Deflection sensitivity: 10 divisions/mA → Current per division $i_d = 0.1$ mA

Galvanometer resistance: $R_g = 60$ Ω

Shunt resistance: $R_s = 1$ Ω

Total divisions: 100

Maximum galvanometer current: $I_g^\text{max} = i_d \times 100 = 0.1 \times 100 = 10$ mA

Maximum current through galvanometer with shunt: $I_\text{max} = I_g^\text{max} \left(1 + \frac{R_g}{R_s}\right)$

$I_\text{max} = 10 \text{ mA} \left(1 + \frac{60}{1}\right) = 10 \text{ mA} \times 61 = 610$ mA

Answer: $I_\text{max} = 610$ mA