Question

A metal wire with circular cross section and length one metre is pulled with tensile force of 1000 N on each side. For the wire to be stretched not more than 0.25 cm, the minimum diameter of the wire required is:
(Young's modulus of the metal $= 10^{11}$ \text{ Pa, take } $\sqrt{\pi} = 1.77$)

• A. \( 1.13 \, \text{mm} \)
• B. \( 2.26 \, \text{mm} \)
• C. \( 4.12 \, \text{mm} \)
• D. \( 3.1 \, \text{mm} \)

Hint:

When dealing with the elongation of a wire, use the formula for Young's modulus and rearrange to solve for the unknown quantity, such as diameter.

Answer 1

The Correct Option is B

We use the formula for the elongation of a metal wire under tensile force: \[ \Delta L = \frac{F L}{A Y} \] Where:
\( \Delta L \) is the elongation of the wire,
\( F \) is the force applied,
\( L \) is the length of the wire,
\( A \) is the cross-sectional area of the wire,
\( Y \) is the Young's modulus of the material.
Given:
\( F = 1000 \, \text{N} \),
\( L = 1 \, \text{m} \),
\( \Delta L = 0.25 \, \text{cm} = 0.0025 \, \text{m} \),
\( Y = 10^{11} \, \text{Pa} \).
For a circular cross section, the area \( A \) is given by \( A = \pi r^2 \), where \( r \) is the radius of the wire.
Rearranging the formula to solve for the diameter: \[ r^2 = \frac{F L}{\Delta L Y \pi} \] Substitute the given values: \[ r^2 = \frac{1000 \times 1}{0.0025 \times 10^{11} \times \pi} \] \[ r^2 = \frac{1000}{7.85 \times 10^{8}} = 1.27 \times 10^{-6} \] \[ r = 1.13 \times 10^{-3} \, \text{m} = 1.13 \, \text{mm} \] Thus, the minimum diameter is: \[ d = 2r = 2 \times 1.13 \, \text{mm} = 2.26 \, \text{mm} \] Thus, the correct answer is: \[ \boxed{2.26 \, \text{mm}} \]