We have a galvanometer of resistance 25 Ω. It is shunted by a 2.5 Ω wire. The part of total current that flows through the galvanometer is given as

(1) $\frac{I}{{I}_{0}}=\frac{1}{11}$

(2) $\frac{I}{{I}_{0}}=\frac{1}{10}$

(3) $\frac{I}{{I}_{0}}=\frac{3}{11}$

(4) $\frac{I}{{I}_{0}}=\frac{4}{11}$

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