N solid metalic spherical balls are melted and reast into a cylindrical rod whose radius is 3 times that of a spherical ball and height is 4 times the radius of a spherical ball. The value of N is

Options

A. `30`
B. `24`
C. `36`
D. `27`

Answer

To solve the problem, we need to find the value of \( N \), which represents the number of solid metallic spherical balls that are melted and reshaped into a cylindrical rod.

### Step-by-Step Solution:

1. **Define the radius of the spherical ball**:
Let the radius of one spherical ball be \( r \).

2. **Determine the dimensions of the cylindrical rod**:
- The radius of the cylindrical rod is given as \( 3 \) times the radius of the spherical ball. Therefore, the radius of the cylinder, \( R \), is:
\[ R = 3r \]
- The height of the cylindrical rod is given as \( 4 \) times the radius of the spherical ball. Therefore, the height of the cylinder, \( H \), is:
\[ H = 4r \]

3. **Calculate the volume of one spherical ball**:
The volume \( V_s \) of one spherical ball is given by the formula:
\[ V_s = \frac{4}{3} \pi r^3 \]

4. **Calculate the total volume of \( N \) spherical balls**:
The total volume of \( N \) spherical balls is:
\[ V_{total} = N \times V_s = N \times \frac{4}{3} \pi r^3 \]

5. **Calculate the volume of the cylindrical rod**:
The volume \( V_c \) of the cylindrical rod is given by the formula:
\[ V_c = \pi R^2 H \]
Substituting \( R \) and \( H \):
\[ V_c = \pi (3r)^2 (4r) = \pi \times 9r^2 \times 4r = 36 \pi r^3 \]

6. **Set the total volume of the spherical balls equal to the volume of the cylindrical rod**:
\[ N \times \frac{4}{3} \pi r^3 = 36 \pi r^3 \]

7. **Cancel \( \pi r^3 \) from both sides** (assuming \( r \neq 0 \)):
\[ N \times \frac{4}{3} = 36 \]

8. **Solve for \( N \)**:
Multiply both sides by \( 3 \):
\[ 4N = 108 \]
Now divide by \( 4 \):
\[ N = \frac{108}{4} = 27 \]

### Conclusion:
The value of \( N \) is \( 27 \). Thus, 27 spherical balls are melted and reshaped into the cylindrical rod.