A galvanometer, having a resistance of $50 \,\Omega$ gives a full scale deflection for a current of $0.05\, A$. The length in meter of a resistance wire of area of cross-section $2.97× 10^{-2} \,cm^2$ 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 = $5 × {10^{ - 7}}\,\Omega m$)
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(c) $\frac{i}{{{i_g}}} = 1 + \frac{G}{S} \Rightarrow \frac{5}{{0.05}} = 1 + \frac{{50}}{S}$
$ \Rightarrow $ $S = \frac{{50}}{{99}} = \frac{{\rho \times l}}{A}$
$ \Rightarrow $ $l = \frac{{50}}{{99}} \times \frac{{2.97 \times {{10}^{ - 2}} \times {{10}^{ - 4}}}}{{5 \times {{10}^{ - 7}}}} = 3\,m$.
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