The diameter of a roller is 2.4 m and its length is 1.68 m. If its rotates 1000 time to level a ground. Find the area of the ground.

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the dimensions of the roller
- The diameter of the roller is given as 2.4 m.
- The length of the roller (height of the cylinder) is given as 1.68 m.

### Step 2: Calculate the radius of the roller
- The radius \( r \) can be calculated from the diameter using the formula:
\[ r = \frac{\text{diameter}}{2} = \frac{2.4 \, \text{m}}{2} = 1.2 \, \text{m} \]

### Step 3: Calculate the curved surface area (CSA) of the roller
- The formula for the curved surface area of a cylinder is:
\[ \text{CSA} = 2 \pi r h \]
- Substituting the values:
\[ \text{CSA} = 2 \times \frac{22}{7} \times 1.2 \, \text{m} \times 1.68 \, \text{m} \]

### Step 4: Simplify the calculation
- First, calculate \( 2 \times 1.2 = 2.4 \).
- Now substitute back:
\[ \text{CSA} = \frac{22}{7} \times 2.4 \times 1.68 \]

### Step 5: Calculate the area of the ground covered by the roller
- The area of the ground covered after 1000 rotations is given by:
\[ \text{Area of ground} = 1000 \times \text{CSA} \]
- Substitute the CSA value we calculated.

### Step 6: Perform the calculations
1. Calculate \( 2.4 \times 1.68 \):
\[ 2.4 \times 1.68 = 4.032 \]
2. Now calculate \( \frac{22}{7} \times 4.032 \):
\[ \frac{22 \times 4.032}{7} = \frac{88.704}{7} \approx 12.672 \, \text{m}^2 \]
3. Finally, multiply by 1000:
\[ \text{Area of ground} = 1000 \times 12.672 \approx 12672 \, \text{m}^2 \]

### Final Answer
The area of the ground covered by the roller after 1000 rotations is **12,672 m²**.

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