A spherical ball of lead has been melted and made in to smaller balls of half the radius of the original one. How many 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 1: Define the radii
Let the radius of the larger ball be \( R \) and the radius of the smaller balls be \( r \). According to the problem, the radius of the smaller balls is half that of the larger ball, so we have:
\[ r = \frac{R}{2} \]

### Step 2: Calculate the volume of the larger ball
The volume \( V_1 \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the volume of the larger ball is:
\[ V_1 = \frac{4}{3} \pi R^3 \]

### Step 3: Calculate the volume of a smaller ball
Using the same volume formula for the smaller balls, we have:
\[ V_2 = \frac{4}{3} \pi r^3 \]
Substituting \( r = \frac{R}{2} \) into this formula gives:
\[ V_2 = \frac{4}{3} \pi \left(\frac{R}{2}\right)^3 = \frac{4}{3} \pi \frac{R^3}{8} = \frac{4}{24} \pi R^3 = \frac{1}{6} \pi R^3 \]

### Step 4: Relate the volumes
Since the larger ball is melted to form \( n \) smaller balls, the total volume of the smaller balls must equal the volume of the larger ball:
\[ V_1 = n \cdot V_2 \]
Substituting the volumes we calculated:
\[ \frac{4}{3} \pi R^3 = n \cdot \frac{1}{6} \pi R^3 \]

### Step 5: Simplify the equation
We can cancel \( \pi R^3 \) from both sides (assuming \( R \neq 0 \)):
\[ \frac{4}{3} = n \cdot \frac{1}{6} \]

### Step 6: Solve for \( n \)
To find \( n \), multiply both sides by 6:
\[ 6 \cdot \frac{4}{3} = n \]
This simplifies to:
\[ n = 8 \]

### Conclusion
Thus, the number of smaller balls that can be made from the larger ball is:
\[ \boxed{8} \]