How many solid lead balls of diameter 4 cm each can be made from a solid lead ball of radius 8 cm ?

Options

A. 64
B. 32
C. 8
D. 26

Answer

To solve the problem of how many solid lead balls of diameter 4 cm can be made from a solid lead ball of radius 8 cm, we will follow these steps:

### Step 1: Find the volume of the larger lead ball.
The formula for the volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
For the larger ball, the radius \( r_1 \) is 8 cm. Therefore, the volume \( V_1 \) is:
\[ V_1 = \frac{4}{3} \pi (8)^3 \]
Calculating \( (8)^3 \):
\[ (8)^3 = 512 \]
Now substituting back into the volume formula:
\[ V_1 = \frac{4}{3} \pi (512) = \frac{2048}{3} \pi \text{ cm}^3 \]

### Step 2: Find the volume of one smaller lead ball.
The diameter of the smaller ball is 4 cm, which means the radius \( r_2 \) is:
\[ r_2 = \frac{4}{2} = 2 \text{ cm} \]
Now, we calculate the volume \( V_2 \) of the smaller ball:
\[ V_2 = \frac{4}{3} \pi (2)^3 \]
Calculating \( (2)^3 \):
\[ (2)^3 = 8 \]
Now substituting back into the volume formula:
\[ V_2 = \frac{4}{3} \pi (8) = \frac{32}{3} \pi \text{ cm}^3 \]

### Step 3: Determine how many smaller balls can be made from the larger ball.
Let \( x \) be the number of smaller balls that can be made. The total volume of the smaller balls must equal the volume of the larger ball:
\[ V_1 = x \cdot V_2 \]
Substituting the volumes we calculated:
\[ \frac{2048}{3} \pi = x \cdot \frac{32}{3} \pi \]
We can cancel \( \frac{3}{\pi} \) from both sides:
\[ 2048 = 32x \]
Now, solving for \( x \):
\[ x = \frac{2048}{32} \]
Calculating \( \frac{2048}{32} \):
\[ x = 64 \]

### Final Answer:
Thus, the number of solid lead balls of diameter 4 cm that can be made from a solid lead ball of radius 8 cm is **64**.

---