When a galvanometer is shunted by a resistance S, its current...

Let the resistance of the galvanometer be $G$ and its initial current capacity be $I_g$.

When shunted by a resistance $S$, the new current capacity becomes $I = n I_g$.

Using the principle of parallel combination, the potential difference across the galvanometer and the shunt are equal:

$I_g G = (I - I_g) S$

$I_g G = (n I_g - I_g) S$

$G = (n - 1) S$

When the galvanometer is shunted by another resistance $S'$, the new current capacity becomes $I' = n' I_g$.

Similarly, we can write:

$I_g G = (I' - I_g) S'$

$I_g G = (n' I_g - I_g) S'$

$G = (n' - 1) S'$

Equating the two expressions for $G$:

$(n - 1) S = (n' - 1) S'$

$n' - 1 = \dfrac{S(n - 1)}{S'}$

$n' = \dfrac{S(n - 1)}{S'} + 1$

$n' = \dfrac{S(n - 1) + S'}{S'}$

Answer: $\dfrac{S(n-1)+S'}{S'}$