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 $2 .97 \times 10^{- 2} {cm}^{2}$ 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 $= 5 \times 10^{- 7} Ω m)$

Question

Infinity Learn · Accepted Answer

iig $$=1$$  $$+$$  GS  ⇒  $$50.05$$ $$=1$$  $$=50S$$⇒ $$S=$$ $$5099$$  $$=$$  ρ × lA  ⇒  $$l=5099$$ ×  $$2.97$$ × $$10$$−$$2$$× $$10$$−$$45$$ × $$10$$−$$7=3$$ m

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