Question:
2.2 dm3 of lead is to be drawn into a cylindrical wire 0.50 cm in diameter. The length of the wire is
(a) 110 m
(b) 112 m
(c) 98 m
(d) 124 m
Solution:
(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:
$\pi(0.25)^{2} \times h=2.2 \times 10 \times 10 \times 10$
$\Rightarrow h=\frac{2200 \times 7}{22 \times 0.0625}$
$=\frac{700}{0.0625}$
$=11200 \mathrm{~cm}$
$=112 \mathrm{~m}$