Question

An aluminum rod [Young's modulus = 7 × 10⁹ (N / m²)] has a breaking strain of 0.2 % what is the minimum cross-sectional area of the rod in order to support a load of 10⁴ Newtons ? 

(A) 1 × 10^–2 m² 

(B) 1.4 × 10^–3 m² 

(C) 3.5 × 10^–3 m² 

(D) 7.1 × 10^–4 m²

Answer 1

Correct option: (D) 7.1 × 10^–4 m²

Explanation:

Y = {(F/A) / (strain)}

Hence

A = {F / (Y × strain)}

Given : F = 10⁴ N

Y = 7 × 10⁹ 

Strain = 0.2%

A = {(10⁴) / (7 × 10⁹ × 0.002)}

= 0.071 × 10^–2 m

A = 7.1 × 10^–4 m²