To solve this problem, we need to use the principle of volume conservation. When the rod is melted and converted into a wire, the volume of the rod remains equal to the volume of the wire. Let's break this down.

Step 1: Volume of the Rod

The rod is cylindrical with:

• Radius = 7 cm
• Length = 150 cm

The volume $V_{rod}$ of a cylinder is given by the formula: $V_{rod} = \pi r^2 h$ where:

• $r$ is the radius of the rod,
• $h$ is the height (or length) of the rod.

Substituting the values: $V_{rod} = \pi (7)^2 \times 150 = \pi \times 49 \times 150 = 7350 \pi \, \text{cm}^3$

Step 2: Volume of the Wire

The wire is also cylindrical with:

• Length = 6 meters = 600 cm (since 1 meter = 100 cm),
• Let the radius of the wire be $r_w$.

The volume $V_{wire}$ of the wire is: $V_{wire} = \pi r_w^2 \times 600$

Step 3: Equating the Volumes

Since the volumes are equal, we set the two expressions equal to each other: $7350 \pi = \pi r_w^2 \times 600$ Canceling $\pi$ from both sides: $7350 = 600 r_w^2$ Now, solve for $r_w^2$: $r_w^2 = \frac{7350}{600} = 12.25$ Taking the square root of both sides: $r_w = \sqrt{12.25} = 3.5 \, \text{cm}$

Step 4: Diameter of the Wire

The diameter of the wire is twice the radius: $\text{Diameter} = 2 \times 3.5 = 7 \, \text{cm}$

Conclusion

The diameter of the wire is 7 cm. Hence, the correct answer is: 02) 7 cm

Would you like more details or have any questions?

Here are five related questions to further explore:

1. What is the volume of a sphere with the same radius as the rod?
2. How would the solution change if the length of the rod was halved?
3. What if the wire was melted and converted back into a new rod of different dimensions?
4. How is the surface area affected during the melting process?
5. What would happen if the wire was stretched instead of melted?

Tip: In volume conversion problems, always check if the units are consistent throughout the problem.