A galvanometer, having a resistance of 50Ω, gives a full sale deflection for a current of 0.05 A. The length in metre of a resistance wire of area of cross-section 2.97x 10-6 cm2 that can be used to convert the galvanometer into an ammeter which can read a maximum of 5 A current is (specific resistance of the wire = 5 x 10-7Ω m)
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a
8
b
6
c
3
d
1.5
answer is C.
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Detailed Solution
Here i=G+SSigor 5=50+RR×0⋅05∴ R=(50/99)Now, R=ρlA or l=RAρ∴ l=(50/99)×2.97×10−65×10−7=3m