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

Options

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

Answer

To solve the problem step by step, we will follow these calculations:

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

### Step 2: Establish Relationships
From the problem statement:
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
Using the first relationship \( r = 2h \), we can express height \( h \) in terms of \( r \):
\[ h = \frac{r}{2} \]

### Step 4: Substitute \( r \) in Terms of \( R \)
From the second relationship, substitute \( r \):
\[ h = \frac{R/2}{2} = \frac{R}{4} \]

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

### Step 6: Calculate the Volume of a Conical Bullet
The volume \( V_b \) of a conical bullet is given by the formula:
\[ V_b = \frac{1}{3} \pi r^2 h \]
Substituting \( r = \frac{R}{2} \) and \( h = \frac{R}{4} \):
\[ V_b = \frac{1}{3} \pi \left(\frac{R}{2}\right)^2 \left(\frac{R}{4}\right) \]
Calculating this:
\[ V_b = \frac{1}{3} \pi \left(\frac{R^2}{4}\right) \left(\frac{R}{4}\right) = \frac{1}{3} \pi \frac{R^3}{16} = \frac{\pi R^3}{48} \]

### Step 7: Set Up the Equation for Number of Bullets
Let \( x \) be the number of bullets formed. The total volume of all bullets is equal to the volume of the sphere:
\[ x \cdot V_b = V \]
Substituting the volumes:
\[ x \cdot \frac{\pi R^3}{48} = \frac{4}{3} \pi R^3 \]

### Step 8: Solve for \( x \)
Cancel \( \pi R^3 \) from both sides:
\[ x \cdot \frac{1}{48} = \frac{4}{3} \]
Multiplying both sides by 48:
\[ x = 48 \cdot \frac{4}{3} = 16 \cdot 4 = 64 \]

### Conclusion
The number of bullets formed is:
\[ \boxed{64} \]