A spherical ball of lead has been melted and made into identical smaller balls with radius equal to half the radius of the original one. How many such balls can be made ?

Answer

To solve the problem of how many smaller balls can be made from a larger spherical ball of lead, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the radius of the original ball**:
Let the radius of the original spherical ball be \( r \).

2. **Calculate the volume of the original ball**:
The formula for the volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, the volume of the original ball is:
\[ V_{original} = \frac{4}{3} \pi r^3 \]

3. **Determine the radius of the smaller balls**:
It is given that the radius of each smaller ball is half of the original radius. Thus, the radius of the smaller ball is:
\[ r_{small} = \frac{r}{2} \]

4. **Calculate the volume of one smaller ball**:
Using the same volume formula for a sphere, the volume of one smaller ball is:
\[ V_{small} = \frac{4}{3} \pi \left(\frac{r}{2}\right)^3 \]
Simplifying this:
\[ V_{small} = \frac{4}{3} \pi \left(\frac{r^3}{8}\right) = \frac{4}{24} \pi r^3 = \frac{1}{6} \pi r^3 \]

5. **Set up the equation for the number of smaller balls**:
Let \( n \) be the number of smaller balls that can be made. Since the volume of the original ball is equal to the total volume of the smaller balls, we can write:
\[ V_{original} = n \times V_{small} \]
Substituting the volumes we calculated:
\[ \frac{4}{3} \pi r^3 = n \times \frac{1}{6} \pi r^3 \]

6. **Cancel out common terms**:
We can cancel \( \pi r^3 \) from both sides (assuming \( r \neq 0 \)):
\[ \frac{4}{3} = n \times \frac{1}{6} \]

7. **Solve for \( n \)**:
To isolate \( n \), multiply both sides by 6:
\[ 6 \times \frac{4}{3} = n \]
Simplifying the left side:
\[ n = 8 \]

### Conclusion:
Thus, the number of smaller balls that can be made from the original ball is \( \boxed{8} \).

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