Skip to content
Login Register

Practice question

MCQ

Question

A steel wire of length 3 m and cross-sectional area 4 × 10-6 m2 is stretched by 0.3 mm. If the Young's modulus of steel is 2 × 1011 N/m2, what is the force applied?

Options

Choose one · Correct answer highlighted

Explanation

Young's modulus: Y = (F L) / (A ΔL). Rearrange: F = (Y A ΔL) / L. Substitute: ΔL = 0.3 × 10-3 m, so F = (2 × 1011 × 4 × 10-6 × 0.3 × 10-3) / 3 = 80 N.

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.

Related blogs

Guides and insights connected to this topic