A 50   Ω resistance is connected to a battery of 5   V . A galvanometer of resistance 100…

Initially current in the circuit is 

$\mathrm{I}=\frac{5}{50}=0.1$

and the current in circuit when ammeter is connected,is $1 \%$ of initial current , i.e.,  ${\mathrm{I}}^{\mathrm{\prime }}=\frac{1}{100}\times 0·1 = 0.099 A$

To increase the range of galvanometer, let resistance $r_s$
 is connected parallel to it and along with $50 \Omega$ is connected in series to ammeter. So, equivalent resistance  of circuit is:

${\mathrm{R}}_{eq}=50+\frac{100 r_{\mathrm{S}}}{100+{\mathrm{r}}_{\mathrm{S}}}$

Now, From equation $V= I^{\text{'}} R_{eq}$

$$\begin{gathered} 5 = 0.099\times \left(50+\frac{100 r_s}{100+r_s}\right) \\ \Rightarrow 50+\frac{100 r_s}{100+r_s} = \frac{5}{0.099} \\ \Rightarrow 50+\frac{100 r_s}{100+r_s} =50.5050 \\ \Rightarrow \frac{100 r_s}{100+r_s} = 50.5050 \\ \Rightarrow r_s = 0.5 \Omega \end{gathered}$$