Question

A galvanometer has a 50 division scale. Battery has no internal resistance. It is found that there is deflection of 40 divisions when r=2400 . Deflection becomes 20 divisions when resistance taken from resistance box is 4900 . Then we can conclude:

Answer 1

I = V/R

Here V is constant and R being changed from resistance box

r = internal resistance of galvanometer

40D = V/(2400 + r)

20D = V/(4900 + r)

Eq 1 / E 2

2 = (4900 + r)/(2400 + r)

=> 4800 + 2r = 4900 + r

=> r = 100

so internal resistance of galvanometer = 100Ω