A moving coil of galvanometer when shunted with 2\, resistance...

Let $G$ be the resistance of the galvanometer and $I_g$ be the current required for full scale deflection.

When the galvanometer is shunted with $S = 2\,\Omega$, it acts as an ammeter of range $I = 500\text{ mA} = 0.5\text{ A}$. The potential difference across the galvanometer and the shunt is equal:

$I_g G = (I - I_g) S$

$I_g G = (0.5 - I_g) \times 2$

$I_g G + 2 I_g = 1$

$I_g = \dfrac{1}{G + 2}$

When a resistance $R = 470\,\Omega$ is connected in series, it acts as a voltmeter of range $V = 10\text{ V}$. The total voltage is the sum of the voltage drops:

$V = I_g (G + R)$

$10 = I_g (G + 470)$

$I_g = \dfrac{10}{G + 470}$

Equating the two expressions for $I_g$:

$\dfrac{1}{G + 2} = \dfrac{10}{G + 470}$

$G + 470 = 10(G + 2)$

$G + 470 = 10G + 20$

$9G = 450$

$G = 50\,\Omega$

Answer: $50$