Copper Rod Drawn Into Wire: Calculating Thickness

This solution details how to determine the thickness of a wire formed by drawing a copper rod. We will use the principle that the volume of the material remains constant throughout the process.

Understanding the Rod and Wire Dimensions

We start with a copper rod and reshape it into a wire. This means the total amount of copper stays the same.

• Original Rod:
  • Diameter ($d_1$) = 1 cm
  • Length ($h_1$) = 8 cm
• Resulting Wire:
  • Length ($h_2$) = 18 m
  • Thickness (Diameter $d_2$) = ? (This is what we need to find)

Volume Calculation Principles

Both the rod and the wire can be modelled as cylinders. The volume ($V$) of a cylinder is given by the formula: $V = \pi r^2 h$ where $r$ is the radius and $h$ is the length.

Since the volume of copper does not change:

Volume of Rod ($V_1$) = Volume of Wire ($V_2$)

Calculating the Rod's Volume ($V_1$)

First, let's find the radius ($r_1$) of the rod:

$r_1 = \frac{\text{Diameter}}{2} = \frac{d_1}{2} = \frac{1 \text{ cm}}{2} = \frac{1}{2} \text{ cm}$

Now, calculate the volume ($V_1$) of the rod:

$V_1 = \pi r_1^2 h_1$

$V_1 = \pi \left(\frac{1}{2} \text{ cm}\right)^2 \times (8 \text{ cm})$

$V_1 = \pi \left(\frac{1}{4} \text{ cm}^2\right) \times (8 \text{ cm})$

$V_1 = \frac{8\pi}{4} \text{ cm}^3 = 2\pi \text{ cm}^3$

Calculating the Wire's Volume ($V_2$)

The length of the wire ($h_2$) is given in meters, so we need to convert it to centimeters for consistency:

$h_2 = 18 \text{ m} = 18 \times 100 \text{ cm} = 1800 \text{ cm}$

Let the unknown diameter of the wire be $d_2$. Its radius ($r_2$) is:

$r_2 = \frac{d_2}{2}$

The volume ($V_2$) of the wire is:

$V_2 = \pi r_2^2 h_2$

$V_2 = \pi \left(\frac{d_2}{2}\right)^2 \times (1800 \text{ cm})$

$V_2 = \pi \frac{d_2^2}{4} \times 1800 \text{ cm}$

Equating Volumes to Find Wire Thickness

Using the principle of volume conservation ($V_1 = V_2$):

$2\pi \text{ cm}^3 = \pi \frac{d_2^2}{4} \times 1800 \text{ cm}$

Now, we solve for $d_2$. Let's simplify the equation:

Divide both sides by $\pi$:

$2 \text{ cm}^3 = \frac{d_2^2}{4} \times 1800 \text{ cm}$

Isolate $d_2^2$:

$d_2^2 = \frac{2 \text{ cm}^3}{\frac{1800}{4} \text{ cm}}$

$d_2^2 = \frac{2 \text{ cm}^3}{450 \text{ cm}}$

$d_2^2 = \frac{1}{225} \text{ cm}^2$

Finally, take the square root to find the diameter $d_2$:

$d_2 = \sqrt{\frac{1}{225} \text{ cm}^2}$

$d_2 = \frac{1}{15} \text{ cm}$

Final Answer

The thickness of the wire is $\frac{1}{15}$ cm.