Question

Two wires are made of the same material and have the same volume. The first wire has cross-sectional area $ A $ and the second wire has cross-sectional area $ 3A $. If the length of the first wire is increased by $ \Delta l $ on applying a force $ F $, how much force is needed to stretch the second wire by the same amount?

• A. \( 4F \)
• B. \( 6F \)
• C. \( 9F \)
• D. \( F \)

Hint:

To stretch a wire by the same amount, the force required is inversely proportional to the cross-sectional area of the wire. A wire with a larger cross-sectional area requires more force to achieve the same extension.

Answer 1

The Correct Option is C

To solve this problem, we begin by understanding the relationship between stress, strain, and Young's Modulus. When stretching a wire:

1. Volume Consistency: Both wires have the same volume.
The formula for volume is \( V = A_1 \cdot L_1 = A_2 \cdot L_2 \), where \( L_1 \) and \( L_2 \) are the lengths of the wires.
For the first wire: \( V = A \cdot L_1 \).
For the second wire: \( V = 3A \cdot L_2 \).
Since volumes are equal: \( A \cdot L_1 = 3A \cdot L_2 \).
Simplifying gives: \( L_2 = \frac{L_1}{3} \).

2. Force Calculation:
The change in length \( \Delta l \) is the same for both wires.
Young's Modulus \( Y \) is given by:
\( Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F \cdot L_1}{A \cdot \Delta l} \).
Rearranging, \( F = \frac{Y \cdot A \cdot \Delta l}{L_1} \).
For the second wire with the same \( \Delta l \):
\( F' = \frac{Y \cdot 3A \cdot \Delta l}{L_2} \).
Since \( L_2 = \frac{L_1}{3} \), substitute to find:
\( F' = \frac{Y \cdot 3A \cdot \Delta l}{\frac{L_1}{3}} = \frac{Y \cdot 9A \cdot \Delta l}{L_1} \).
This shows \( F' = 9F \).

Conclusion: The force needed to stretch the second wire by the same amount \( \Delta l \) is \( 9F \). Therefore, the correct answer is \( \boxed{9F} \).

Answer 2

The two wires are made of the same material and have the same volume. This implies that the product of cross-sectional area and length is constant for both wires. If the first wire has cross-sectional area \(A\) and length \(L_1\), and the second wire has cross-sectional area \(3A\) and length \(L_2\), we have: \[A \cdot L_1 = 3A \cdot L_2\] Solving for \(L_2\): \[L_2 = \frac{L_1}{3}\] The change in length when a force is applied to a wire is given by the equation: \[\Delta l = \frac{F \cdot L}{A \cdot Y}\] where \(F\) is the force, \(L\) is the original length, \(A\) is the cross-sectional area, and \(Y\) is the Young's modulus of the material (same for both wires since they are of the same material). For the first wire: \[\Delta l = \frac{F \cdot L_1}{A \cdot Y}\] Rearranging for the force \(F_1\): \[F_1 = \frac{\Delta l \cdot A \cdot Y}{L_1}\] For the second wire, let \(F_2\) be the force required to achieve the same elongation \(\Delta l\): \[\Delta l = \frac{F_2 \cdot L_2}{3A \cdot Y}\] Rearranging for \(F_2\): \[F_2 = \frac{\Delta l \cdot 3A \cdot Y}{L_2}\] Since \(L_2 = \frac{L_1}{3}\), substitute it into the equation: \[F_2 = \frac{\Delta l \cdot 3A \cdot Y}{\frac{L_1}{3}} = \frac{\Delta l \cdot 3A \cdot Y \cdot 3}{L_1} = \frac{\Delta l \cdot 9A \cdot Y}{L_1}\] Comparing the equations for \(F_1\) and \(F_2\): \[F_2 = 9 \cdot \frac{\Delta l \cdot A \cdot Y}{L_1} = 9F_1\] Therefore, the force needed to stretch the second wire by the same amount \(\Delta l\) is \(9F\). The correct answer is \(9F\).