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

Given: Galvanometer resistance $$G = 72 \;\Omega$$ and shunt resistance $$S = 8 \;\Omega$$.

When a shunt is connected in parallel with the galvanometer, the total current $$I$$ splits between the galvanometer and the shunt.

Since the galvanometer and shunt are in parallel, the voltage across both is the same:

$$I_g \times G = I_s \times S$$

where $$I_g$$ is the current through the galvanometer and $$I_s$$ is the current through the shunt.

Also, $$I = I_g + I_s$$, so $$I_s = I - I_g$$.

$$I_g \times 72 = (I - I_g) \times 8$$

$$72 I_g = 8I - 8I_g$$

$$80 I_g = 8I$$

$$\frac{I_g}{I} = \frac{8}{80} = \frac{1}{10}$$

The percentage of total current through the galvanometer:

$$\frac{I_g}{I} \times 100 = \frac{1}{10} \times 100 = 10\%$$

The correct answer is Option B.