On melting a solid sphere of lead of radius 8 cm , find the number of spherical balls of radius 1 cm that can be mode?

Answer

To solve the problem of finding the number of spherical balls of radius 1 cm that can be made from melting a solid sphere of lead with a radius of 8 cm, we will follow these steps:

### Step 1: Calculate the volume of the larger sphere
The formula for the volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
For the larger sphere with a radius \( r = 8 \) cm:
\[ V_{\text{large}} = \frac{4}{3} \pi (8)^3 \]

### Step 2: Calculate \( 8^3 \)
Calculating \( 8^3 \):
\[ 8^3 = 8 \times 8 \times 8 = 512 \]
Thus, the volume of the larger sphere becomes:
\[ V_{\text{large}} = \frac{4}{3} \pi (512) = \frac{2048}{3} \pi \text{ cm}^3 \]

### Step 3: Calculate the volume of the smaller sphere
Now, we calculate the volume of the smaller sphere with a radius \( r = 1 \) cm:
\[ V_{\text{small}} = \frac{4}{3} \pi (1)^3 = \frac{4}{3} \pi \text{ cm}^3 \]

### Step 4: Find the number of smaller spheres
Let \( n \) be the number of smaller spheres that can be made. The total volume of the smaller spheres must equal the volume of the larger sphere:
\[ n \cdot V_{\text{small}} = V_{\text{large}} \]
Substituting the volumes we calculated:
\[ n \cdot \frac{4}{3} \pi = \frac{2048}{3} \pi \]

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

### Conclusion
Thus, the number of spherical balls of radius 1 cm that can be made from melting the larger sphere is:
\[ \boxed{512} \]

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