Question

A copper rod of diameter 1 cm and 8 cm length is stretched to form a wire of uniform thickness with length 18 m. Find the thickness of the wire.

Hint:

When a solid is stretched into a wire, the volume remains constant. Use the volume formula for a cylinder to find the new dimensions.

Answer 1

The volume of the copper rod is the same as the volume of the wire formed by stretching it. The volume of a cylinder is given by the formula: \[ V = \pi r^2 h, \] where \( r \) is the radius and \( h \) is the height (or length).
Step 1: The radius of the copper rod is \( r = \frac{1}{2} = 0.5 \, \text{cm} \) and the length of the rod is \( h = 8 \, \text{cm} \). The volume of the copper rod is: \[ V_{\text{rod}} = \pi (0.5)^2 \times 8 = \pi \times 0.25 \times 8 = 2 \pi \, \text{cm}^3. \]
Step 2: When the copper rod is stretched into a wire, the volume remains the same. Let the radius of the wire be \( r_w \) and the length of the wire be \( h_w = 18 \, \text{m} = 1800 \, \text{cm} \). The volume of the wire is: \[ V_{\text{wire}} = \pi r_w^2 \times 1800. \] Since the volume of the rod is equal to the volume of the wire, we have: \[ 2 \pi = \pi r_w^2 \times 1800. \] Cancel \( \pi \) from both sides: \[ 2 = 1800 r_w^2 \quad \Rightarrow \quad r_w^2 = \frac{2}{1800} = \frac{1}{900}. \]
Step 3: Solve for \( r_w \): \[ r_w = \frac{1}{30} \, \text{cm}. \]
Step 4: The thickness of the wire is twice the radius: \[ \text{Thickness} = 2r_w = \frac{2}{30} = \frac{1}{15} \, \text{cm}. \]
Conclusion: The thickness of the wire is \( \frac{1}{15} \, \text{cm} \).