A galvanometer gives full scale deflection with 0 · 006   A current. By connecting it to a 4990…

Given a galvanometer of resistance (of resistance $R_G$) get converted into a voltmeter of range $0-30V$ , when a resistance of $R_S = 4990 \Omega$ is connected in series of it and give full scale of deflection with current $I_G = 0·006 A$.

Than from equation ${\mathrm{I}}_{\mathrm{G}}=\frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{S}}+R_{\mathrm{G}}}$
$0.006=\frac{30}{4990+\mathrm{G}}$ $\Rightarrow R_G = 10 \Omega$

Now this galvanometer is connected with a shunt resistance to convert into ammeter of range $0- 1·5 A$

Given ${\mathrm{I}}_{\mathrm{G}}=0.006$, so current passing shunt will be 
${\mathrm{I}}_{\mathrm{S}}=I-I_G=1.5-0·006 = 1.494 A$
Since ${\mathrm{R}}_{\mathrm{G}}$ and ${\mathrm{R}}_{\mathrm{S}}$ are in parallel, ${\mathrm{I}}_{\mathrm{G}}{\mathrm{R}}_{\mathrm{G}}={\mathrm{I}}_{\mathrm{s}}{\mathrm{R}}_{\mathrm{S}}$
$0.006\mathrm{R}=1.494\left(\frac{2\mathrm{n}}{249}\right)$ ;   $\because$given $R_S=\left(\frac{2\mathrm{n}}{249}\right)$  
$\therefore \mathrm{n}=5$