How many identical spherical bullets can be made out of a solid metallic cube, if the edge of the cube is 11 times the diameter of the bullet ?

Options

A. 317
B. 635
C. 1271
D. 2541

Answer

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

### Step 1: Define the Variables
Let the diameter of the bullet be \( d \). According to the problem, the edge of the cube is \( 11d \).

### Step 2: Calculate the Volume of the Cube
The volume \( V \) of a cube with edge length \( a \) is given by the formula:
\[ V = a^3 \]
Here, the edge length of the cube is \( 11d \), so the volume of the cube is:
\[ V_{\text{cube}} = (11d)^3 = 1331d^3 \]

### Step 3: Calculate the Volume of One Bullet
The volume \( V \) of a sphere (bullet) with diameter \( d \) is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Since the radius \( r \) is half of the diameter, we have:
\[ r = \frac{d}{2} \]
Substituting this into the volume formula gives:
\[ V_{\text{bullet}} = \frac{4}{3} \pi \left(\frac{d}{2}\right)^3 = \frac{4}{3} \pi \frac{d^3}{8} = \frac{1}{6} \pi d^3 \]

### Step 4: Set Up the Equation for the Number of Bullets
Let \( n \) be the number of bullets that can be formed from the cube. The total volume of the bullets must equal the volume of the cube:
\[ n \cdot V_{\text{bullet}} = V_{\text{cube}} \]
Substituting the volumes we calculated:
\[ n \cdot \frac{1}{6} \pi d^3 = 1331d^3 \]

### Step 5: Solve for \( n \)
To find \( n \), we can rearrange the equation:
\[ n = \frac{1331d^3}{\frac{1}{6} \pi d^3} \]
The \( d^3 \) cancels out:
\[ n = 1331 \cdot \frac{6}{\pi} \]
Substituting \( \pi \approx \frac{22}{7} \):
\[ n = 1331 \cdot \frac{6 \cdot 7}{22} = 1331 \cdot \frac{42}{22} = 1331 \cdot \frac{21}{11} \]
Calculating this gives:
\[ n = 2541 \]

### Conclusion
The number of identical spherical bullets that can be made from the solid metallic cube is \( \boxed{2541} \).