A galvanometer can be converted into a voltmeter of certain range...

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

When a resistance of $990\ \Omega$ is connected in series, the range of the voltmeter is $V$.

$V = I_g (G + 990)$

When a resistance of $480\ \Omega$ is connected in series, the range is halved, i.e., $\dfrac{V}{2}$.

$\dfrac{V}{2} = I_g (G + 480)$

Dividing the first equation by the second equation:

$\dfrac{V}{\dfrac{V}{2}} = \dfrac{I_g (G + 990)}{I_g (G + 480)}$

$2 = \dfrac{G + 990}{G + 480}$

$2(G + 480) = G + 990$

$2G + 960 = G + 990$

$G = 30\ \Omega$

Answer: $30\ \Omega$