How many lead balls, each of radius 1 cm, can be made from a sphere of radius 8 cm ?

Answer

To solve the problem of how many lead balls, each with a radius of 1 cm, can be made from a sphere of radius 8 cm, we will follow these steps:

### Step 1: Understand the volumes involved
We need to find the volume of the larger sphere (with radius 8 cm) and the volume of the smaller lead balls (with radius 1 cm). The volume of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]

### Step 2: Calculate the volume of the larger sphere
Using the formula for the volume of a sphere, we calculate the volume of the larger sphere:
\[ V_{\text{large}} = \frac{4}{3} \pi (R^3) = \frac{4}{3} \pi (8^3) \]
Calculating \(8^3\):
\[ 8^3 = 512 \]
So,
\[ V_{\text{large}} = \frac{4}{3} \pi (512) = \frac{2048}{3} \pi \text{ cm}^3 \]

### Step 3: Calculate the volume of one lead ball
Now, we calculate the volume of one lead ball:
\[ V_{\text{small}} = \frac{4}{3} \pi (r^3) = \frac{4}{3} \pi (1^3) \]
Calculating \(1^3\):
\[ 1^3 = 1 \]
So,
\[ V_{\text{small}} = \frac{4}{3} \pi (1) = \frac{4}{3} \pi \text{ cm}^3 \]

### Step 4: Determine the number of lead balls
To find the number of lead balls \(n\), we set the volume of the larger sphere equal to the total volume of the smaller lead balls:
\[ n \cdot V_{\text{small}} = V_{\text{large}} \]
Substituting the volumes we calculated:
\[ n \cdot \frac{4}{3} \pi = \frac{2048}{3} \pi \]

### Step 5: Solve for \(n\)
We can cancel \(\frac{4}{3} \pi\) from both sides:
\[ n = \frac{2048}{4} = 512 \]

### Final Answer
Thus, the number of lead balls that can be made from the larger sphere is:
\[ \boxed{512} \]