Question

When a wire of resistance R is stretched to twice its original length?

Answer 1

When a wire is stretched to twice its original length, its resistance changes due to alterations in its dimensions. Here's a step-by-step explanation of how this works:

### 1. **Initial Resistance**

Let's start with the original wire. Assume:
- **Initial resistance**: \( R \)
- **Initial length**: \( L \)
- **Initial cross-sectional area**: \( A \)

The resistance \( R \) of the wire is given by:
\[ R = \rho \frac{L}{A} \]
where \( \rho \) is the resistivity of the material.

### 2. **Stretching the Wire**

When the wire is stretched to twice its original length, the new length \( L' \) is:
\[ L' = 2L \]

### 3. **Volume Conservation**

Assuming the wire is stretched uniformly, the volume of the wire is conserved. The volume \( V \) before stretching is:
\[ V = A \times L \]

After stretching, the volume \( V' \) should be the same, so:
\[ V' = A' \times L' \]
where \( A' \) is the new cross-sectional area.

Since \( V = V' \):
\[ A \times L = A' \times L' \]
\[ A \times L = A' \times (2L) \]
\[ A' = \frac{A}{2} \]

The new cross-sectional area \( A' \) is half of the original cross-sectional area.

### 4. **New Resistance**

The resistance \( R' \) of the wire after stretching is:
\[ R' = \rho \frac{L'}{A'} \]

Substitute \( L' = 2L \) and \( A' = \frac{A}{2} \):
\[ R' = \rho \frac{2L}{\frac{A}{2}} \]
\[ R' = \rho \frac{2L \times 2}{A} \]
\[ R' = 4 \rho \frac{L}{A} \]
\[ R' = 4R \]

### **Conclusion**

When a wire is stretched to twice its original length, its resistance increases by a factor of 4.

Answer 2

When a wire of resistance \( R \) is stretched to twice its original length, several factors affect its resistance. Let’s analyze the changes step by step.

### Key Relationships:
1. **Resistance Formula**:
   The resistance \( R \) of a wire is given by the formula:
   \[
   R = \rho \frac{L}{A}
   \]
   where:
   - \( \rho \) is the resistivity of the material (a constant for a given material),
   - \( L \) is the length of the wire,
   - \( A \) is the cross-sectional area of the wire.

2. **Change in Length**:
   When the wire is stretched to twice its original length, the new length becomes \( 2L \) (i.e., twice the original length).

3. **Effect on Cross-Sectional Area**:
   The volume of the wire remains constant during stretching, which means:
   \[
   \text{Initial volume} = \text{Final volume}
   \]
   Volume is given by \( V = A \times L \). So, if the length doubles, the cross-sectional area \( A \) must decrease to maintain the same volume.

   Let the original cross-sectional area be \( A_0 \) and the original length be \( L_0 \). After stretching, the new length is \( 2L_0 \) and the new cross-sectional area is \( A_1 \). Since the volume remains constant:
   \[
   A_0 \times L_0 = A_1 \times (2L_0)
   \]
   Solving for \( A_1 \):
   \[
   A_1 = \frac{A_0}{2}
   \]
   So, the cross-sectional area becomes half of the original area.

### New Resistance:
Now, we can calculate the new resistance \( R_1 \). Using the resistance formula again, for the new length \( 2L_0 \) and the new area \( A_1 = \frac{A_0}{2} \):
   \[
   R_1 = \rho \frac{2L_0}{A_1} = \rho \frac{2L_0}{\frac{A_0}{2}} = \rho \frac{4L_0}{A_0}
   \]
Comparing this to the original resistance \( R_0 = \rho \frac{L_0}{A_0} \), we see:
   \[
   R_1 = 4R_0
   \]
   
### Conclusion:
When a wire is stretched to twice its original length, its resistance becomes **four times** the original resistance. This increase occurs because the wire becomes longer and thinner, both of which contribute to an increase in resistance.