A $72 \mathrm{\Omega }$ galvanometer is shunted by a resistance of $8 \mathrm{\Omega }$. The percentage of the total current which passes through the galvanometer is 

Let $G$ be resistance of galvanometer and $I\text{'}$ be the current passing through the galvanometer produces full scale deflection. If $I$ is the maximum current, and since, $G$ and $S$ are in parallel so, the potential difference between them will be same.

Then,

 $$\begin{gathered} I^{\text{'}}G=I_{2}S \\ \Rightarrow 72I^{\text{'}}=8I_2 \\ \Rightarrow I_2=9I^{\text{'}} \\ \Rightarrow I=I^{\text{'}}+I_2=10I^{\text{'}} \end{gathered}$$

So the percentage of total current passes through galvanometer is 

$\frac{I^{\text{'}}}{I}\times 100$$=\frac{I^{\text{'}}}{10I^{\text{'}}}\times 100=10\%$