Question

How many times resistance will increase if length of conducting wire becomes twice when stretched?

Answer 1

To understand how the resistance of a conducting wire changes when its length is increased, we can use the formula for resistance in terms of its physical properties:

\[
R = \rho \frac{L}{A}
\]

where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material (which remains constant for a given material),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

### Initial Resistance

Let’s assume the initial length of the wire is \( L \) and its cross-sectional area is \( A \). The initial resistance \( R_1 \) can be expressed as:

\[
R_1 = \rho \frac{L}{A}
\]

### Stretching the Wire

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

\[
L' = 2L
\]

Now, let's consider the change in cross-sectional area. When a wire is stretched, its volume remains constant (assuming no material is added or removed). The initial volume \( V \) of the wire can be expressed as:

\[
V = L \times A
\]

After stretching, the volume \( V' \) remains the same, but the length has changed to \( L' \) and the new cross-sectional area \( A' \) will be different. We can express this as:

\[
V' = L' \times A' = 2L \times A'
\]

Setting the initial volume equal to the final volume gives:

\[
L \times A = 2L \times A'
\]

From this, we can solve for the new cross-sectional area \( A' \):

\[
A' = \frac{A}{2}
\]

### New Resistance

Now we can find the new resistance \( R_2 \) using the new length and new area:

\[
R_2 = \rho \frac{L'}{A'} = \rho \frac{2L}{\frac{A}{2}} = \rho \frac{2L \cdot 2}{A} = \rho \frac{4L}{A}
\]

### Ratio of Resistances

Now we can find the ratio of the new resistance \( R_2 \) to the original resistance \( R_1 \):

\[
\frac{R_2}{R_1} = \frac{\rho \frac{4L}{A}}{\rho \frac{L}{A}} = \frac{4L}{L} = 4
\]

### Conclusion

Thus, when the length of the conducting wire is stretched to twice its original length, its resistance will increase by a factor of **4**.

Answer 2

To determine how many times the resistance of a conducting wire increases when its length is doubled, we need to consider the relationship between resistance, length, and the material's properties.

### Resistance of a Wire

The resistance \( R \) of a wire is given by the formula:

\[ R = \rho \frac{L}{A} \]

where:
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

### Effect of Stretching the Wire

When a wire is stretched to twice its original length, several changes occur:

1. **Length**: The new length \( L' \) is \( 2L \).

2. **Volume Conservation**: Assuming the wire is stretched uniformly, its volume remains constant. The original volume \( V \) is:

   \[ V = L \times A \]

   When stretched, the new volume \( V' \) is:

   \[ V' = L' \times A' = 2L \times A' \]

   Since the volume is constant:

   \[ L \times A = 2L \times A' \]

   Solving for the new cross-sectional area \( A' \):

   \[ A' = \frac{A}{2} \]

### New Resistance

Substitute the new length \( L' \) and cross-sectional area \( A' \) into the resistance formula:

\[
R' = \rho \frac{L'}{A'}
\]

Substitute \( L' = 2L \) and \( A' = \frac{A}{2} \):

\[
R' = \rho \frac{2L}{\frac{A}{2}} = \rho \frac{2L \times 2}{A} = 4 \rho \frac{L}{A}
\]

Thus:

\[
R' = 4R
\]

### Conclusion

The resistance of the wire will increase by a factor of 4 when the length of the wire is doubled by stretching it.