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

Current through the galvanometer,I=350  +2950  =  10−3  ACurrent for 30 divisions  =  10−3  ACurrent for 20 divisions = 10−330  ×  20=23  ×  10−3  AFor the same deflection to obtain for 20 division, let resistance added be R.∴       23   ×   10−3  =  350+Ror     R=4450   Ω
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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