A cylindrical roller is of length 2 m and diameter 1.4 m. Find the area it will level in2500 revolutions.

1. Break Down the Problem

To solve the problem, we need to determine:

• The circumference of the cylindrical roller to find out how far it travels in one revolution.
• The lateral surface area of the cylinder, which gives the area it will level in one revolution.
• Finally, multiply the area leveled in one revolution by the total number of revolutions (2500) to find the total area leveled.

2. Relevant Concepts

• Circumference of a cylinder: $C = \pi \times d$, where $d$ is the diameter.
• Lateral surface area of a cylinder: $A = C \times h$, where $h$ is the height (or length) of the cylinder.

3. Analysis and Detail

1. Calculate the circumference:

   • The diameter $d = 1.4 \, \text{m}$.
   • Circumference $C = \pi \times d = \pi \times 1.4$.

   $C \approx 3.14 \times 1.4 \approx 4.398 \, \text{m}$

2. Calculate the lateral surface area for one revolution:

   • The length $h = 2 \, \text{m}$.
   • Area per revolution $A = C \times h = 4.398 \times 2$.

   $A \approx 4.398 \times 2 \approx 8.796 \, \text{m}^2$

3. Calculate the total area leveled in 2500 revolutions:

   • Total area $A_{total} = A \times \text{number of revolutions} = 8.796 \times 2500$.

   $A_{total} \approx 8.796 \times 2500 \approx 21990 \, \text{m}^2$

4. Verify and Summarize

The calculations were verified and followed a logical progression. The circumference was computed correctly, leading to the lateral surface area and finally the total area leveled.

Final Answer

The cylindrical roller will level approximately $21990 \, \text{m}^2$ in 2500 revolutions.