Question

If a copper rod whose diameter is 1 cm and length 8 cm is melted and recasted into another rod whose length is 18 cm, then find the width of the rod.

Hint:

To solve this problem, remember that the volume of the rod remains constant after recasting, and use the formula for the volume of a cylinder to set up an equation.

Answer 1

Step 1: Volume of the original rod.
The original rod is in the shape of a cylinder. The formula for the volume of a cylinder is: \[ V = \pi r^2 h \]
where:
- \( r \) is the radius of the rod, and
- \( h \) is the height (or length) of the rod.
The diameter of the rod is 1 cm, so the radius \( r = \frac{1}{2} = 0.5 \, \text{cm} \). The length of the rod is \( h = 8 \, \text{cm} \). Substituting these values into the volume formula: \[ V = \pi (0.5)^2 \times 8 = \pi \times 0.25 \times 8 = 2 \pi \, \text{cubic cm.} \]
Step 2: Volume of the new rod.
The volume of the copper remains the same after recasting. The volume of the new rod is also \( 2 \pi \, \text{cubic cm.} \), but the length of the new rod is given as 18 cm. Let \( r' \) be the radius (width) of the new rod. Using the formula for the volume of a cylinder again: \[ V = \pi (r')^2 \times 18 \] Equating the volumes of the original and new rods: \[ 2 \pi = \pi (r')^2 \times 18 \] Dividing both sides by \( \pi \): \[ 2 = (r')^2 \times 18 \] Solving for \( r' \): \[ r'^2 = \frac{2}{18} = \frac{1}{9} \] \[ r' = \frac{1}{3} \, \text{cm.} \] Thus, the diameter of the new rod is \( 2r' = 2 \times \frac{1}{3} = \frac{2}{3} \, \text{cm.} \)
Conclusion:
The width (diameter) of the new rod is \( \boxed{\frac{2}{3}} \, \text{cm.} \)