A 72Ω galvanometer is shunted by a resistance of 8Ω . The...

Let $G$ be resistance of galvanometer and $I '$ 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,

${I}^{'}G = {I}_{2}S ⇒ 72{I}^{'} = 8{I}_{2} ⇒ {I}_{2} = 9{I}^{'} ⇒ I = {I}^{'} + {I}_{2} = 10{I}^{'}$

So the percentage of total current passes through galvanometer is

$\dfrac{{I}^{'}}{I} \times 100$$= \dfrac{{I}^{'}}{10{I}^{'}} \times 100 = 10$