A cylindrical copper rod is reformed to twice its original length with no change in volume. The resistance between its ends before the change was R. Now its resistance will be :

Options

A. 8R
B. 6R
C. 4R
D. 2R

Answer

To solve the problem step by step, we need to analyze the changes in the cylindrical copper rod's dimensions and how they affect its resistance.

### Step 1: Understand the relationship between resistance, length, and area
The resistance \( R \) of a cylindrical conductor is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the conductor,
- \( A \) is the cross-sectional area.

### Step 2: Define the original parameters
Let:
- The original length of the rod be \( L \).
- The original cross-sectional area be \( A \).
- The original resistance be \( R \).

From the formula, we can express the original resistance as:
\[ R = \frac{\rho L}{A} \]

### Step 3: Determine the new length and area after reforming
The problem states that the rod is reformed to twice its original length. Therefore, the new length \( L' \) is:
\[ L' = 2L \]

Since the volume of the rod remains constant, we can express the volume before and after reforming:
\[ \text{Original Volume} = A \cdot L \]
\[ \text{New Volume} = A' \cdot L' = A' \cdot (2L) \]
Setting these two volumes equal gives:
\[ A \cdot L = A' \cdot (2L) \]
We can cancel \( L \) from both sides (assuming \( L \neq 0 \)):
\[ A = 2A' \]
Thus, the new cross-sectional area \( A' \) is:
\[ A' = \frac{A}{2} \]

### Step 4: Calculate the new resistance
Now we can find the new resistance \( R' \) using the new dimensions:
\[ R' = \frac{\rho L'}{A'} \]
Substituting \( L' \) and \( A' \):
\[ R' = \frac{\rho (2L)}{\frac{A}{2}} = \frac{2\rho L \cdot 2}{A} = \frac{4\rho L}{A} \]

### Step 5: Relate the new resistance to the original resistance
We know from the original resistance formula that:
\[ R = \frac{\rho L}{A} \]
Thus, we can express \( R' \) in terms of \( R \):
\[ R' = 4 \cdot \frac{\rho L}{A} = 4R \]

### Conclusion
The new resistance \( R' \) after reforming the rod is:
\[ R' = 4R \]