(b) As $m = z\ l\ t = z\left( \ \frac{V}{R} \right)t\ \ ie,\ m \propto Vt$
$\therefore\ $ $\frac{m_{2}}{m_{1}} = \frac{V_{2}t_{2}}{V_{1}t_{1}}$
or ${\ m}_{2} = \frac{V_{2}t_{2}}{V_{1}t_{2}} \times m_{1} = \frac{6 \times 45 \times 2}{12 \times 30} =$1.5g.