A metallic sphere is melted and moulded to form conical shaped bullets. If radius of the bullet is a twice of its height and radius of bullet is half of the radius of the metallic sphere, then what is the numbers of bullet formed?

Options

A. 32
B. 16
C. 128
D. 64

Answer

To solve the problem step by step, we need to find out how many conical bullets can be formed from a metallic sphere when certain conditions are given.

### Step 1: Define the variables
Let:
- The radius of the bullet = \( r \)
- The height of the bullet = \( h \)
- The radius of the metallic sphere = \( R \)

### Step 2: Establish relationships from the problem statement
According to the problem:
1. The radius of the bullet is twice its height:
\[ r = 2h \]
2. The radius of the bullet is half of the radius of the metallic sphere:
\[ r = \frac{R}{2} \]

### Step 3: Express height in terms of radius
From the first relationship, we can express height \( h \) in terms of radius \( r \):
\[ h = \frac{r}{2} \]

### Step 4: Calculate the volume of the bullet
The volume \( V_b \) of a cone is given by the formula:
\[ V_b = \frac{1}{3} \pi r^2 h \]
Substituting \( h \) with \( \frac{r}{2} \):
\[ V_b = \frac{1}{3} \pi r^2 \left(\frac{r}{2}\right) = \frac{1}{6} \pi r^3 \]

### Step 5: Calculate the volume of the metallic sphere
The volume \( V_s \) of a sphere is given by the formula:
\[ V_s = \frac{4}{3} \pi R^3 \]

### Step 6: Substitute \( R \) in terms of \( r \)
Since \( r = \frac{R}{2} \), we can express \( R \) as:
\[ R = 2r \]
Now substituting this into the volume of the sphere:
\[ V_s = \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi (8r^3) = \frac{32}{3} \pi r^3 \]

### Step 7: Set up the equation for the number of bullets
Let \( n \) be the number of bullets formed. The total volume of the bullets must equal the volume of the sphere:
\[ n \cdot V_b = V_s \]
Substituting the volumes we calculated:
\[ n \cdot \left(\frac{1}{6} \pi r^3\right) = \frac{32}{3} \pi r^3 \]

### Step 8: Solve for \( n \)
Dividing both sides by \( \pi r^3 \):
\[ n \cdot \frac{1}{6} = \frac{32}{3} \]
Multiplying both sides by 6:
\[ n = 6 \cdot \frac{32}{3} = 64 \]

### Conclusion
The number of bullets formed is \( \boxed{64} \).