Question

A steel rod has a radius of 10 mm and a length of 1.0 m. A force stretches it along its length and produces a strain of 0.16%. Young's modulus of the steel is 2.0 x10¹¹N/m² . What is the magnitude of the force stretching the rod?

Answer 1

Answer: 100kN

Answer 2

As we know that according to hooke's law, the stress in an object is directly proportional to strain developed in the object.

Hooke's law- stress ∝ strain

After reducing the proportionality sign, the given relation can be represented as

σ = ∈ x Y

Where, \(\sigma = \frac{F}{A} = Stress\)

∈ = \(\frac{\Delta l}{l} = Strain\)

Y = Young’s modulus of elasticity

\(\Delta l\) = change in length

F =  Applied force

A = Cross sectional area

Given data in question as following

The equation of hooke's law can be written as

\(\frac{F}{A} = \frac{Y \Delta l} {l}\)

After putting the values in this equation, we get

F = 2 x 10¹¹ x 3.14 x 10⁻⁴ x 0.16 x 10⁻²

F = 100 kN