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A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of 4 × 10⁻⁴ A passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V, it should be connected to a resistance of

Asked in JEE Main 12th Jan 2nd Shift 2019 · Conversion to ammeter and voltmeter

Answer: (1) 200 ohm

Step-by-step solution

I_g = 25 × 4 × 10⁻⁴ = 10⁻² A.

Total resistance = (2.5)/(10⁻²) = 250 Ω.

Series resistance = 250 - 50 = 200 Ω.

Why the other options are wrong

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