A steel rod has a radius of 20 mm and a length of 2.0 m. A force of 62.8 kN stretches it along its length. Young's modulus of steel is 2.0 × 10¹¹ N / m². The longitudinal strain produced in the wire is _____________ × 10⁻⁵

Solution

Physics sir · 3 steps

Strain is just stress divided by Young's modulus, and the rod's length plays no part here.

1. Cross-section area. Convert the radius to metres first: r = 20 mm = 0.02 m. Then A = π r² = 3.14 × 4 × 10⁻⁴ = 1.256 × 10⁻³ m².
2. Find the stress. Stress = F/A = (62.8 × 10³)/(1.256 × 10⁻³) = 5 × 10⁷ N m⁻².

   Strain = Stress/Y = (5 × 10⁷)/(2.0 × 10¹¹) = 2.5 × 10⁻⁴

3. So. 2.5 × 10⁻⁴ = 25 × 10⁻⁵, so the blank is 25.

The common slip: 2.5. Stopping at 2.5 × 10⁻⁴ and entering 2.5 ignores the × 10⁻⁵ in the blank. Rewrite it as 25 × 10⁻⁵ before answering.

The idea to keep. Strain needs only force, area and Y. The length matters only if you are asked for the extension, Δ L = strain × L. Always match your answer to the power of ten in the blank.

Answer from the NTA final answer key. Solution by PrepWiser.