2.2 dm3 of lead is to be drawn into a cylindrical wire 0.50 cm in diameter.

(b) 112 m

Volume of the lead = volume of the cylindrical wire
Suppose that h m is the length of the wire.

As $2.2 \mathrm{dm}^{3}$ of lead is to be drawn into a cylindrical wire of diameter $0.50 \mathrm{~cm}$, we have:

$\pi(0.25)^{2} \times h=2.2 \times 10 \times 10 \times 10$

$\Rightarrow h=\frac{2200 \times 7}{22 \times 0.0625}$

$=\frac{700}{0.0625}$

$=11200 \mathrm{~cm}$

$=112 \mathrm{~m}$