Question

An iron rod of length 2m and cross-sectional area of 50mm² stretched by 0.5mm, when a mass of 250 kg is hung from its lower end. Young’s modulus of iron rod is

• A. \( 19.6 \times 10^{20} \, \text{N/m}^2 \)
• B. \( 19.6 \times 10^{18} \, \text{N/m}^2 \)
• C. \( 19.6 \times 10^{10} \, \text{N/m}^2 \)
• D. \( 19.6 \times 10^{15} \, \text{N/m}^2 \)

Hint:

Young’s modulus is a measure of the stiffness of a material and is calculated as the ratio of stress to strain.

Answer 1

The Correct Option is C

Step 1: Young’s Modulus Formula.
Young’s modulus \( Y \) is given by: \[ Y = \frac{F L}{A \Delta L} \] where \( F \) is the force, \( L \) is the length of the rod, \( A \) is the cross-sectional area, and \( \Delta L \) is the elongation. The force \( F \) is the weight of the mass, i.e., \( F = mg \).
Step 2: Calculation.
Substitute the given values to calculate the Young’s modulus. After solving, we find that \( Y = 19.6 \times 10^{10} \, \text{N/m}^2 \).
Step 3: Conclusion.
The correct answer is (C), \( 19.6 \times 10^{10} \, \text{N/m}^2 \).