Question

A cylindrical rod made of aluminum has length 1 meter and diameter of 10 cm. The rod is subjected to a tensile force of 100 kN. The elongation in the rod is:

• A. \( 0.81 \times 10^{-4} \, \text{m} \)
• B. \( 2 \times 10^{-4} \, \text{m} \)
• C. \( 0.2 \times 10^{-4} \, \text{m} \)
• D. \( 1.81 \times 10^{-4} \, \text{m} \)

Hint:

Use the formula for elongation \( \Delta L = \frac{F L}{A Y} \), where \( F \) is the force, \( L \) is the length, \( A \) is the cross-sectional area, and \( Y \) is Young’s modulus.

Answer 1

The Correct Option is D

Using Hooke’s law and the formula for elongation in a material subjected to tensile force, we calculate the elongation in the rod. The result is \( 1.81 \times 10^{-4} \, \text{m} \). Thus, the correct answer is option (4).