Question

When a uniform resistance wire is stretched till its length becomes four times does it resistance increase or decrease?

Answer 1

When a uniform resistance wire is stretched so that its length becomes four times its original length, its resistance increases. Here's why:

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. **Effect of Stretching**: When the wire is stretched to four times its original length, the length \( L \) becomes \( 4L \). To maintain the same volume, the cross-sectional area \( A \) of the wire must decrease. Since the volume of the wire is the product of its length and cross-sectional area, stretching it while keeping the volume constant means the area \( A \) will be reduced.

3. **Cross-Sectional Area**: If the original length is \( L \) and the original cross-sectional area is \( A \), then after stretching, the new length is \( 4L \) and the new cross-sectional area \( A' \) can be found by:
   \[
   A \times L = A' \times 4L
   \]
   Solving for \( A' \) gives:
   \[
   A' = \frac{A}{4}
   \]

4. **New Resistance**: The new resistance \( R' \) after stretching is:
   \[
   R' = \rho \frac{4L}{A'}
   \]
   Substituting \( A' = \frac{A}{4} \):
   \[
   R' = \rho \frac{4L}{\frac{A}{4}} = \rho \frac{4L \times 4}{A} = 16 \rho \frac{L}{A} = 16R
   \]
   So, the resistance increases by a factor of 16.

Therefore, the resistance of the wire increases by a factor of 16 when its length is stretched to four times its original length.

Answer 2

When a uniform resistance wire is stretched so that its length becomes four times longer, its resistance increases. Here's why:

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, and \( A \) is the cross-sectional area.

When the wire is stretched to four times its original length, the length \( L \) increases by a factor of 4. For the volume of the wire to remain constant, the cross-sectional area \( A \) decreases. The new cross-sectional area can be determined from the original area \( A_0 \) and the new length \( L \) as follows:

\[ A_{\text{new}} = \frac{A_0}{4} \]

So, the new resistance \( R_{\text{new}} \) can be calculated by:

\[ R_{\text{new}} = \rho \frac{4L_0}{A_0 / 4} = 16 \rho \frac{L_0}{A_0} = 16 R_0 \]

This means the resistance increases by a factor of 16 when the wire's length is stretched to four times its original length.