Question

A uniform metallic wire is elongated by 0.04 m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force will be ________ cm.

Hint:

$\Delta L \propto \frac{L}{d^2}$. Doubling length increases elongation $(\times 2)$, but doubling diameter decreases it $(\times 1/4)$, resulting in a net factor of $1/2$.

Answer 1

Correct Answer: 2

To solve this problem, we must understand how elongation is affected by the dimensions of the wire. The elongation (\( \Delta L \)) of a wire when a force \( F \) is applied can be described by Hooke's Law:

\( \Delta L = \frac{FL}{AE} \) 

where \( L \) is the original length, \( A \) is the cross-sectional area, and \( E \) is the Young's modulus of the material.

Original Wire:

The elongation given is 0.04 m with force \( F \). Let's analyze:

• \( \Delta L_1 = \frac{FL}{A_1E} = 0.04 \, \text{m} \)

Modified Wire:

• Length \( L' = 2L \)
• Diameter is doubled, so new area \( A' = \pi \left(\frac{2d}{2}\right)^2 = 4A_1 \)

The new elongation \( \Delta L' \) is:

\( \Delta L' = \frac{F(2L)}{4A_1E} = \frac{2FL}{4A_1E} = \frac{FL}{2A_1E} \)

Substituting the known elongation for the original wire:

\( \Delta L' = \frac{1}{2} \Delta L_1 = \frac{1}{2} \times 0.04 \, \text{m} = 0.02 \, \text{m} \)

Convert 0.02 m to centimeters:

\( 0.02 \, \text{m} = 2 \, \text{cm} \)

This result is within the expected range [2, 2].

Conclusion: The elongation of the wire when both length and diameter are doubled, keeping the force constant, is 2 cm.