1. Given data

• Full scale deflection current of galvanometer: $I_g = 0.006\ \text{A}$
• When connected with a resistance of $4990\,\Omega$, it becomes a voltmeter of range $0$ to $30\,\text{V}$.
• When connected with a resistance of $\dfrac{2n}{249}\,\Omega$, it becomes an ammeter of range $0$ to $1.5\,\text{A}$.

We first find the galvanometer resistance $G$.

2. Finding galvanometer resistance using voltmeter condition

To convert a galvanometer into a voltmeter of range $V = 30\,\text{V}$, a high resistance is connected in series.

So total resistance required is: $R_{\text{total}} = \frac{V}{I_g} = \frac{30}{0.006} = 5000\,\Omega$

Given series resistance is $4990\,\Omega$, therefore galvanometer resistance is: $G = 5000 - 4990 = 10\,\Omega$

3. Finding shunt resistance for ammeter conversion

To convert galvanometer into an ammeter of range $I = 1.5\,\text{A}$, a small resistance $S$ is connected in parallel with the galvanometer.

At full scale:

• Current through galvanometer = $I_g = 0.006\,\text{A}$
• Current through shunt = $I_s = I - I_g = 1.5 - 0.006 = 1.494\,\text{A}$

Since galvanometer and shunt are in parallel, voltage across them is same: $I_g G = I_s S$

Thus, $S = \frac{I_g G}{I_s} = \frac{0.006 \times 10}{1.494}$

$S = \frac{0.06}{1.494} = \frac{60}{1494} = \frac{10}{249}\,\Omega$

Given shunt resistance is: $S = \frac{2n}{249}\,\Omega$

So, $\frac{2n}{249} = \frac{10}{249}$

Hence, $2n = 10 \Rightarrow n = 5$

4. Final answer

$\boxed{5}$

The derived answer matches the stored correct answer.