Question

A force \( F \) applied on the wire of radius \( r \) and length \( L \) and change in the length of the wire is \( l \). If the same force \( F \) is applied on the wire of the same material and radius \( 4r \) and length \( 4L \), then change in length of the other wire is

• A. \( \frac{l}{4} \)
• B. \( 2l \)
• C. \( \frac{l}{2} \)
• D. \( 4l \)

Hint:

When calculating elongation, remember that the elongation is inversely proportional to the cross-sectional area and directly proportional to the length.

Answer 1

The Correct Option is A

We use the formula for the elongation of a wire under a force, which is given by: \[ \Delta L = \frac{F L}{A Y} \] where: - \( F \) is the applied force,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire,
- \( Y \) is the Young’s Modulus of the material.
The change in length of the first wire is \( \Delta L_1 = \frac{F L}{\pi r^2 Y} \), and the change in length of the second wire is: \[ \Delta L_2 = \frac{F (4L)}{\pi (4r)^2 Y} = \frac{4F L}{\pi \cdot 16r^2 Y} = \frac{1}{4} \cdot \Delta L_1 \] Thus, the change in length of the second wire is \( \frac{l}{4} \), where \( l \) is the change in length of the first wire. Therefore, the correct answer is (A).