The value of the shunt resistance that allows $10 \%$ of the main current through the galvanometer of $99 \Omega$ is

The galvanometer with resistance $G = 99\,\Omega$ carries $I_g = 0.10 I$ of the main current $I$. The shunt resistance $S$ carries the remaining current $I_s = 0.90 I$.

Since the galvanometer and shunt are in parallel, the voltage across both must be equal:

$I_g G = I_s S$

Substituting the current expressions:

$(0.10 I) G = (0.90 I) S$

Canceling $I$ and solving for $S$:

$S = \frac{0.10}{0.90} G = \frac{1}{9} \times 99 = 11\,\Omega$

The shunt resistance required is $11\,\Omega$, corresponding to option B.