SSC JE Electrical Engineering 2010 Paper

Given the measurement of Galvanometer resistance = 3a and shunting resistor = 30Ω
Ammeter = 30||30
Resistance = 15
$\mathit{m}=\frac{{\mathit{I}}_{\mathrm{Ext}}}{{\mathit{I}}_{\mathit{m}}}=2$${\mathit{I}}_{\mathrm{Ext}}=2{\mathit{I}}_{\mathit{m}}$
${\mathit{I}}_{\mathrm{Ext}}={\mathit{I}}_{\mathit{m}}+{\mathit{I}}_{\mathit{s}}$
${\mathit{I}}_{\mathit{d}}=2{\mathit{I}}_{\mathit{m}}-{\mathit{I}}_{\mathit{n}}={\mathit{I}}_{\mathit{m}}$
${\mathit{I}}_{\mathit{m}}\times 15=\mathit{I}\mathit{S}\times {\mathit{R}}_{\mathit{S}}$
${\mathit{I}}_{\mathit{m}}15={\mathit{I}}_{\mathit{m}}{\mathit{R}}_{\mathit{S}}$
${\mathit{R}}_{\mathit{S}}=15\mathit{\Omega }$
So the additional shunt should be connected across it to double the range be 15Ω.