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Q.

A resistance R1 is connected in the left gap of a metre bridge and a resistance of 9Ω is connected the right gap in the metre bridge. The null point divides the bridge wire in the ratio 4:3. Now another resistance x Ω is connected in series with R1 in the left gap of te metre bridge and this time the null point divides the bridge wire in the ratio 4:1. Then the value of ‘x’ is

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a

18 Ω

b

24 Ω

c

16 Ω

d

8 Ω

answer is B.

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Detailed Solution

R1R2=43.....(1)and R1+xR2=41.....(2)∴ R1R2+xR2=4⇒43+xR2=4⇒xR2=4−43−83∴ x=83R2=83×9 Ω=24 Ω
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