Question

IR
29. A galvanometer of resistance 50. n is connected 29
to a battery of 3V along with a resistance of
2950 o in series shows full-scale deflection of
30 divisions. The additional series resistance
required
to reduce the deflection to 20
divisions is
(A) 15002
(B) 4440
(C) 7400
(D) 2950​

Answer 1

Explanation:

A galvanometer of resistance 50 Ω is connected to a battery along with a resistance 2950 Ω in series.

so, Req =

= 50 + 2950 = 3000 Ω

now, current through galvanometer, I = 3v/3000 = 10^-3 A

current for 30 divisions = I = 10^-3 A

⇒current for 1 division = I/30 = 1/30 × 10^-3 A

⇒current for 20 divisions = 20I/30 = 20/30 × 10^-3 A = 2 × 10^-3/3 A

now, equivalent resistance will be , Req = 3v/(2 × 10^-3/3) = 9000/2 = 4500 Ω

as it is given, resistance of galvanometer is 50 Ω

so, resistance in series should be , R = Req -

= 4500 - 50 = 4450 Ω

So Answer Should be B

Hope it helps

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