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A galvanometer having a coil resistance of 100 Ω gives a full-scale deflection when a current of 1 mA is passed through it. The value of the resistance which can convert this galvanometer into ammeter giving a full-scale deflection for a current of 10 A is 

Answer:

Explanation:

 In parallel, current distributes in inverse ratio of resistance

 Hence,  $\frac{l-l_{g}}{l_{g}}=\frac{G}{S} \Rightarrow S=\frac{GI_{g}}{I-I_{g}}$

As I_g is very small, hence    $S=\frac{GI_{g}}{I}$

 $b= \frac{(100)(1\times 10^{-3})}{10}=0.01$ Ω