A galvanometer has a coil of resistance 100 ohm and gives a full scale deflection for 30mA current.  If it is work as a voltmeter of 30 volt range, the resistance(in ohm) required to be added will be 

$\left(a\right)$: Here, resistance of galvanometer, 
$G=100\Omega$ 
Current for ul scale deflection, $l_g=30mA$ 
 $=30\times {10}^{-3}A$
Range of voltmeter, $V=30V$ 
To convert the galvanometer into an voltmeter of a given range, a resistance R is connected in series with it as shown in the figure.  
From figure. $V=l_g\left(G+R\right)$ 
Or $R=\frac{V}{l_g}-G$ 
$=\frac{30}{30\times {10}^{-3}}-100\Omega =1000-100=900\Omega$