A metal resistance R is stretched to double its length. Its new resistance will be ………….. .

Answer

To solve the problem of finding the new resistance when a metal resistance \( R \) is stretched to double its length, we can follow these steps:

### Step-by-Step Solution:

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

2. **Identify the initial conditions**:
Let the initial length of the resistance be \( L \) and the initial area of cross-section be \( A \). Therefore, the initial resistance is:
\[ R = \frac{\rho L}{A} \]

3. **Determine the new length**:
When the resistance is stretched to double its length, the new length \( L' \) becomes:
\[ L' = 2L \]

4. **Conserve the volume**:
The volume of the conductor remains constant during stretching. The volume \( V \) can be expressed as:
\[ V = A \times L \]
After stretching, the new volume \( V' \) is:
\[ V' = A' \times L' = A' \times (2L) \]
Setting the volumes equal gives:
\[ A \times L = A' \times (2L) \]
Simplifying this, we find:
\[ A' = \frac{A}{2} \]

5. **Calculate the new resistance**:
Now, substituting the new length and area into the resistance formula:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (2L)}{\frac{A}{2}} = \frac{2\rho L \cdot 2}{A} = \frac{4\rho L}{A} \]
This shows that:
\[ R' = 4R \]

### Final Answer:
The new resistance when the metal resistance \( R \) is stretched to double its length will be \( 4R \).