A galvanometer of resistance $90\ \Omega$ is shunted by a resistance of $10\ \Omega$. The fraction of main current that passes through shunt is

Let the main current be $I$, the current through the shunt be $I_s$, and the current through the galvanometer be $I_g$.

Since the galvanometer and the shunt are connected in parallel, the potential difference across them is equal.

$I_s S = I_g G$

Using $I = I_s + I_g$, we can write $I_g = I - I_s$.

$I_s S = (I - I_s) G$

$I_s (S + G) = I G$

$\dfrac{I_s}{I} = \dfrac{G}{S + G}$

Given $G = 90\ \Omega$ and $S = 10\ \Omega$:

$\dfrac{I_s}{I} = \dfrac{90}{10 + 90} = \dfrac{90}{100} = \dfrac{9}{10}$

Answer: $9 / 10$