A long solid cylinder is radiating power. It is remolded into a number of smaller cylinders, each of which has the same length as original cylinder. Each small cylinder has the same temperature as the original cylinder. The total radiant power emitted by the pieces is twice that emitted by the original cylinder. How many smaller cylinders are there? (Neglect the energy emitted by the flat faces of cylinder.)

Options

A. 3
B. 4
C. 5
D. 6

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Power Emission
The power emitted by a body due to radiation is given by the formula:
\[ P = E \cdot A \cdot \sigma \cdot T^4 \]
where:
- \( P \) is the power emitted,
- \( E \) is the emissivity (which is constant for both the original and smaller cylinders),
- \( A \) is the surface area,
- \( \sigma \) is the Stefan-Boltzmann constant (also constant),
- \( T \) is the absolute temperature (same for both).

Since the temperature and emissivity are the same for both the original and smaller cylinders, we can focus on the surface area.

### Step 2: Calculate the Surface Area of the Original Cylinder
The surface area \( A_1 \) of the original cylinder (neglecting the flat faces) is given by:
\[ A_1 = 2 \pi R L \]
where \( R \) is the radius and \( L \) is the length of the cylinder.

### Step 3: Calculate the Surface Area of the Smaller Cylinders
Let’s assume we mold the original cylinder into \( n \) smaller cylinders, each with radius \( r \) and the same length \( L \). The surface area \( A_2 \) of one smaller cylinder is:
\[ A_2 = 2 \pi r L \]
Thus, the total surface area of all \( n \) smaller cylinders is:
\[ A_{total} = n \cdot A_2 = n \cdot (2 \pi r L) = 2 \pi n r L \]

### Step 4: Relate the Power Emitted
According to the problem, the total radiant power emitted by the smaller cylinders is twice that emitted by the original cylinder:
\[ A_{total} = 2 \cdot A_1 \]
Substituting the areas we found:
\[ 2 \pi n r L = 2 \cdot (2 \pi R L) \]
We can simplify this by canceling out common terms:
\[ n r = 2 R \]

### Step 5: Use Volume Conservation
The volume of the original cylinder must equal the total volume of the smaller cylinders since the material is conserved. The volume \( V_1 \) of the original cylinder is:
\[ V_1 = \pi R^2 L \]
The volume \( V_2 \) of \( n \) smaller cylinders is:
\[ V_2 = n \cdot \pi r^2 L \]
Setting these equal gives:
\[ \pi R^2 L = n \cdot \pi r^2 L \]
Canceling out \( \pi L \) (assuming \( L \neq 0 \)):
\[ R^2 = n r^2 \]

### Step 6: Solve the Equations
From the equations \( n r = 2 R \) and \( R^2 = n r^2 \), we can substitute \( R \) from the first equation into the second:
1. From \( n r = 2 R \), we get \( R = \frac{n r}{2} \).
2. Substitute \( R \) into \( R^2 = n r^2 \):
\[ \left(\frac{n r}{2}\right)^2 = n r^2 \]
\[ \frac{n^2 r^2}{4} = n r^2 \]
Dividing both sides by \( r^2 \) (assuming \( r \neq 0 \)):
\[ \frac{n^2}{4} = n \]
Rearranging gives:
\[ n^2 - 4n = 0 \]
Factoring out \( n \):
\[ n(n - 4) = 0 \]
Thus, \( n = 0 \) or \( n = 4 \).

### Conclusion
Since \( n = 0 \) does not make sense in this context, we conclude that:
\[ n = 4 \]
Therefore, the number of smaller cylinders is **4**.