Question

A wire of length 'L' and radius 'r' is clamped rigidly at one end. When the other end of the wire is pulled by a force f, its length increases by 'l'. Another wire of same material of length '2L' and radius '2r' is pulled by a force '2f'. Then the increase in its length will be

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

Answer 1

The Correct Option is A

To solve this problem, we need to use the concept of elasticity and specifically the formula for elongation due to a tensile force. The elongation (\( \Delta L \)) of a material subjected to a force is given by:

\(\Delta L = \frac{FL}{A \cdot Y}\)

where:

• \( F \) is the applied force
• \( L \) is the original length of the wire
• \( A \) is the cross-sectional area of the wire
• \( Y \) is the Young's modulus of the material

For the first wire:

• Length = \( L \)
• Radius = \( r \)
• Applied Force = \( f \)
• Elongation = \( l \)

The cross-sectional area \( A \) for the first wire is \( \pi r^2 \).

The elongation is given by:

\(l = \frac{fL}{\pi r^2 Y}\)

For the second wire:

• Length = \( 2L \)
• Radius = \( 2r \)
• Applied Force = \( 2f \)

The cross-sectional area \( A \) for the second wire is \( \pi (2r)^2 = 4\pi r^2 \).

The elongation \( \Delta L \) for the second wire is given by:

\(\Delta L = \frac{(2f)(2L)}{4\pi r^2 Y}\)

Simplifying the expression:

\(\Delta L = \frac{4fL}{4\pi r^2 Y} = \frac{fL}{\pi r^2 Y} = l\)

Thus, the elongation in the second wire is the same as in the first wire, which is \( l \). Therefore, the correct answer is l.