Question

A copper wire of diameter 6 mm is evenly wrapped on a cylinder of length 18 cm and diameter 49 cm to cover its whole surface. Find the length and the volume of the wire. If the density of the copper be 8.8 g per cm³, then find the weight of the wire.

Answer 1

$\mathrm{We}\mathrm{have},\mathrm{Diameter}\mathrm{of}\mathrm{the}\mathrm{coppe}\mathrm{wire},d=6\mathrm{mm}=0.6\mathrm{cm},\mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{copper}\mathrm{wire},r=\frac{0.6}{2}=0.3\mathrm{cm},\mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{cylinder},H=18\mathrm{cm},\mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{cylinder},R=\frac{49}{2}\mathrm{cm}\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{rotations}\mathrm{of}\mathrm{the}\mathrm{wire}\mathrm{on}\mathrm{the}\mathrm{cylinder}=\frac{\mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{cylinder},H}{\mathrm{Diameter}\mathrm{of}\mathrm{the}\mathrm{copper}\mathrm{wire},d}=\frac{18}{0.6}=30\mathrm{The}\mathrm{circumference}\mathrm{of}\mathrm{the}\mathrm{base}\mathrm{of}\mathrm{the}\mathrm{cylinder}=2\pi R=2\times \frac{22}{7}\times \frac{49}{2}=154\mathrm{cm}\mathrm{So},\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{wire},h=30\times 154=4620\mathrm{cm}=46.2m\mathrm{Now},\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{wire}=\pi r^{2}h=\frac{22}{7}\times 0.3\times 0.3\times 4620=1306.8{\mathrm{cm}}^3\mathrm{Also},\mathrm{the}\mathrm{weight}\mathrm{of}\mathrm{the}\mathrm{wire}=\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{wire}\times \mathrm{Density}\mathrm{of}\mathrm{the}\mathrm{wire}=1306.8\times 8.8=11499.84g\approx 11.5\mathrm{kg}$