A cylindrical rod of iron whose height is eight times its radius is melted and cast into spherical balls each of half the radius of the cylinder. The number of such spherical balls is

Options

A. a) 12
B. b) 16
C. c) 24
D. d) 48

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the radius of the cylindrical rod
Let the radius of the cylindrical rod be \( r \). According to the problem, the height of the rod is \( 8r \).

### Step 2: Calculate the volume of the cylindrical rod
The formula for the volume \( V \) of a cylinder is given by:
\[ V = \pi r^2 h \]
Substituting the height \( h = 8r \):
\[ V = \pi r^2 (8r) = 8\pi r^3 \]

### Step 3: Define the radius of the spherical balls
The radius of each spherical ball is given as half of the radius of the cylindrical rod. Therefore, the radius of the spherical ball is:
\[ \text{Radius of sphere} = \frac{r}{2} \]

### Step 4: Calculate the volume of one spherical ball
The formula for the volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting the radius of the sphere \( \frac{r}{2} \):
\[ V = \frac{4}{3} \pi \left(\frac{r}{2}\right)^3 = \frac{4}{3} \pi \left(\frac{r^3}{8}\right) = \frac{4}{24} \pi r^3 = \frac{1}{6} \pi r^3 \]

### Step 5: Determine the number of spherical balls
Let \( n \) be the number of spherical balls formed. The total volume of the spherical balls must equal the volume of the cylindrical rod:
\[ n \times \text{Volume of one sphere} = \text{Volume of cylinder} \]
Substituting the volumes we calculated:
\[ n \times \frac{1}{6} \pi r^3 = 8 \pi r^3 \]

### Step 6: Solve for \( n \)
To find \( n \), we can divide both sides by \( \pi r^3 \):
\[ n \times \frac{1}{6} = 8 \]
Multiplying both sides by 6:
\[ n = 8 \times 6 = 48 \]

### Conclusion
The number of spherical balls that can be made by melting the cylindrical rod is \( n = 48 \).

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