Question

A copper wire has a radius of 4.1 mm. When forces of a certain equal magnitude but opposite directions are applied to the ends of the wire, the wire stretches by 5.0×10−3 of its original length. What is the tensile stress on the wire? Young's modulus for copper is 11×1010Pa.

Answer 1

Answer:

tensile stress = 55 x 10⁷ Pa

Explanation:

given,

radius of the copper wire = 4.1 mm = 0.0041 m

the wire stretches = 5.0×10⁻³ l

   l = original length

young's modulus of copper = 11 x 10¹⁰ Pa

strain = $\dfrac{\Delta L}{L} = 5 \times 10^{-3}$

$young\ modulus = \dfrac{stress}{strain}$

$young\ modulus = \dfrac{stress}{strain}$

tensile stress = young's modulus x strain

tensile stress = 11 x 10¹⁰ x 5.0×10⁻³

tensile stress = 55 x 10⁷ Pa

The tensile stress on the wire is equal to 55 x 10⁷ Pa