A galvanometer, having a resistance of 50 Ω gives a full scale deflection for a current of 0.05 A. The length in metre of a resistance wire of area of cross-section 2.97x 10-2 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 wire = 5 x 10-7 Ωm)
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a
8
b
6
c
3
d
1.5
answer is C.
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Detailed Solution
We know that i−igS=igG∴ i=(G+S)SigTherefore 5=50+RR×0⋅05or R=2⋅54⋅95Further R=ρlA or l=ARρor l=2⋅97×10−6×(2⋅5)5×10−7×4⋅95=3m