A galvanometer of resistance 50 Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

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5050 Ω

5550 Ω

6050 Ω

4450 Ω

answer is D.

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Detailed Solution

initial current i=33000=10−3A10−3A→30divisions20div→10−3×23A For current 23×10−3 resistance supposed to be, 10−3×23=3R+50R+50=92×103=4500R=4450Ω

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