In a galvanometer there is deflection of 10 divisions per mA Thje internal resistance of the galvanometer is 60 ohm - Physics - Current Electricity

Question

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?

Ankur · Accepted Answer

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