Question

A copper wire has a radius of 3.5 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:

The tensile stress on the wire is 550 MPa.

Explanation:

Given;

Radius of copper wire, R = 3.5 mm

extension of the copper wire, e =  5.0×10⁻³ L

L is the original length of the copper wire,

Young's modulus for copper, Y =  11×10¹⁰Pa.

Young's modulus, Y is given as the ratio of tensile stress to tensile strain, measured in the same unit as Young's modulus.

$Y =\frac{Tensile \ stress}{Tensile \ strain} \\\\Tensile \ stress = Y*Tensile \ strain\\\\But, Tensile \ strain = \frac{extension}{original \ Length} = \frac{5.0*10^{-3} L}{L} = 5.0*10^{-3}\\\\Tensile \ stress = 11*10^{10} *5.0*10^{-3} \ = 550*10^6 \ Pa$

Therefore, the tensile stress on the wire is 550 MPa.