Question

A galvanometer of resistance 50Ω is connected to a battery of 3V along with a resistance of 2950Ω in series shown, full-scale deflection of 30 divisions. The additional series resistance required to reduce the deflection to 20 divisions is

A
4440Ω
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B
1500Ω
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C
7400Ω
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D
2950Ω
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Solution

The correct option is C 1500ΩThe current flowing through the galvanometer in series with 2950Ω is I=VR=32950+50=0.001AThus, the galvanometer deflects by 30 divisions for a current of 0.001A.Let the added resistance in series be R.The current flowing through the galvanometer now is 32950+50+RThe galvanometer deflection is 20 divisions.We know that, deflection is proportional to the current in the galvanometer.Thus, 2030=33000+R0.001⇒3000+R3000=32⇒R=1500Ω

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