Question

The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young’s modulus, respectively, are 8 × 10⁸ N m^–2 and 2 × 10¹¹ N m^–2, is :

• A. 4 mm
• B. 0.4 mm
• C. 40 mm
• D. 8 mm

Answer 1

The Correct Option is A

Step 1: Use the Formula for Elongation 

The elongation of a wire is given by:

$$ \Delta L = \frac{\sigma L}{Y} $$

\(\Delta L\) = Elongation of the wire

\(\sigma\) = Stress (Elastic limit)

L = Length of the wire

Y = Young’s modulus

Step 2: Substitute the Given Values

Given:

\(\sigma = 8 \times 10^8 \text{ N/m}^2\)

\(L = 1 \text{ m}\)

\(Y = 2 \times 10^{11} \text{ N/m}^2\)

Substituting these values into the equation:

$$ \Delta L = \frac{(8 \times 10^8) \times 1}{2 \times 10^{11}} $$

Step 3: Convert the Result

Solving the equation:

$$ \Delta L = 4 \times 10^{-3} \text{ m} $$

Converting to millimeters:

$$ \Delta L = 4 \text{ mm} $$

Conclusion

The elongation of the wire is 4 mm.