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Q.

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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a

5050 Ω

b

5550 Ω

c

6050 Ω

d

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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