Problem 16 - University Physics with Modern Physics

Question

A ductile metal wire has resistance \(R .\) What will be the resistance of this wire in terms of \(R\) if it is stretched to three times its original length, assuming that the density and resistivity of the material do not change when the wire is stretched? (Hint: The amount of metal does not change, so stretching out the wire will affect its cross-sectional area.)

Step-by-Step Solution

Answer

The resistance becomes 9R.

Step 1: Understanding the Relationship

The resistance of a wire depends on its resistivity \(\rho\), length \(L\), and cross-sectional area \(A\) as given by the formula: \[ R = \frac{\rho L}{A}. \] When the wire is stretched to three times its length, both the length and cross-sectional area will be affected.

Step 2: Analyzing Volume Constancy

The volume of the wire, which is the product of its cross-sectional area \(A\) and length \(L\), remains constant during stretching. Therefore, \(A_1L_1 = A_2L_2\). If the original length \(L\) is stretched to \(3L\), then:\[ AL = A_2(3L). \] Thus, the new cross-sectional area \(A_2 = \frac{A}{3}.\)

Step 3: Calculating New Resistance

Substitute the new length and area into the resistance formula. The new resistance \(R'\) is given by:\[ R' = \frac{\rho (3L)}{A/3} = \frac{3\rho L}{A} \times 3 = 9 \times \frac{\rho L}{A} = 9R. \]

Step 4: Conclusion

By stretching the wire to three times its original length, the resistance of the wire increases to nine times its original resistance.

Key Concepts

Resistivity

When we talk about resistivity, we are referring to a specific property of a material that tells us how much it resists the flow of electric charge. This is an intrinsic property, meaning it doesn’t depend on the shape or size of the material, but rather on its composition. Think of it like a special fingerprint for how a material handles electricity.
Resistivity is represented by the Greek letter \( \rho \) (rho), and is part of the resistance formula \( R = \frac{\rho L}{A} \).
Here's how resistivity fits in:

• It gives us a measure of how strongly a material opposes the flow of electrical current.
• Higher resistivity means a material is a better insulator, while lower resistivity means it's a better conductor.

What's important to note in this exercise is that the resistivity \( \rho \) remains unchanged when the wire is stretched. This means that despite changes in length and cross-sectional area, the resistive property of the material stays the same.

Cross-Sectional Area

The cross-sectional area of a wire is crucial in determining its resistance because it effectively sets how wide the pathway for electricity is. Imagine trying to fit a large crowd through a narrow doorway; a narrow passage would make it much harder for everyone to get through quickly.
In our context, the cross-sectional area \( A \) plays a significant role in the formula \( R = \frac{\rho L}{A} \).
Here's why it matters:

• Larger cross-sectional areas provide more room for electric charge to flow, reducing resistance.
• Smaller cross-sectional areas increase resistance, as they limit the space for charge movement.

For the wire in this exercise, when it is stretched, the new cross-sectional area \( A_2 \) is reduced to a third of its original size \( A \). This decrease occurs because the overall volume of the wire remains unchanged, leading us to the principle of volume constancy.

Volume Constancy

The concept of volume constancy is fundamental in materials that are being reshaped or extended. For the wire exercise, the volume of the wire remains the same before and after stretching. This is because the wire is simply being elongated without adding or removing any material.
Here's how volume constancy works:

• The original volume \( V \) is calculated as \( V = A_1L_1 \).
• When stretched to three times its length, the volume equation becomes \( A_2 \times 3L = A_1 \times L \).
• As a result, the new cross-sectional area \( A_2 \) becomes \( \frac{A_1}{3} \).

This preserves the material's volume while changing its shape. Thus, although the length of the wire changes and its resistance increases, the total amount of material and hence its volume, remains the same. By applying this principle, we can infer that the wire's new resistance becomes nine times its original resistance because length affects resistance significantly, as shown in our calculations.