The number of spherical bullets that can be made out of a solid cube of lead whose edge measures 88 cm, each bullet being 4 cm in diameter, is

Options

A. 25000
B. 25440
C. 20328
D. 25140

Answer

To solve the problem of determining how many spherical bullets can be made from a solid cube of lead, we will follow these steps:

### Step 1: Calculate the Volume of the Cube
The volume \( V \) of a cube is given by the formula:
\[ V = \text{side}^3 \]
Given that the edge of the cube is 88 cm, we can calculate the volume:
\[ V = 88^3 = 88 \times 88 \times 88 \]

### Step 2: Calculate the Volume of One Bullet
The volume \( V \) of a sphere (bullet) is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
The diameter of each bullet is 4 cm, so the radius \( r \) is:
\[ r = \frac{4}{2} = 2 \text{ cm} \]
Now, substituting the radius into the volume formula:
\[ V = \frac{4}{3} \pi (2)^3 = \frac{4}{3} \pi \times 8 = \frac{32}{3} \pi \text{ cm}^3 \]

### Step 3: Set Up the Equation for the Number of Bullets
Let \( n \) be the number of bullets. The total volume of the bullets must equal the volume of the cube:
\[ \text{Volume of cube} = n \times \text{Volume of one bullet} \]
This gives us the equation:
\[ 88^3 = n \times \frac{32}{3} \pi \]

### Step 4: Solve for \( n \)
Rearranging the equation to solve for \( n \):
\[ n = \frac{88^3}{\frac{32}{3} \pi} = \frac{88^3 \times 3}{32 \pi} \]

### Step 5: Calculate \( n \)
Now we will compute \( 88^3 \):
\[ 88^3 = 681472 \]
Substituting this value into the equation for \( n \):
\[ n = \frac{681472 \times 3}{32 \pi} \]
Calculating \( \frac{681472 \times 3}{32} \):
\[ n = \frac{2044416}{32 \pi} \]
Now, using \( \pi \approx 3.14 \):
\[ n \approx \frac{2044416}{32 \times 3.14} \approx \frac{2044416}{100.48} \approx 20340.5 \]
Since \( n \) must be a whole number, we round down to the nearest whole number:
\[ n \approx 20340 \]

### Final Answer
The number of spherical bullets that can be made from the solid cube of lead is approximately **20340**.