In a galvanometer there is deflection of 10 divisions per mA.Thje internal resistance of the galvanometer is 60 ohm.If a shunt of 2.5 ohm is connected to the galvanometer and there are 50 divisions in all,on the scale of galvanometer,what maximum current can this galvanometer read?

MY DOUBT IS:(If wrong plz correct it and explain) Since there are 50 divisions in all in the galvanometer,then maximum current that galvanometer would read is Ig=50*1/10=5 mA .But the answer is given 125 mA.How?

As the galvanometer has 50 divisions, current required to produce full scale deflection is :
Ig = 110×50 mA = 5 mARg = 60 ΩRs = 2.5 ΩLet I be the maximum current that the galvanometer can read.Then, Ig = RgRg +  Rs × Ior, I = (Rg +  Rs)Rs× Ig  = (60 + 2.5)2.5×5 = 125 mA

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