Question

A weight exerts force of 120N on steel wire of cross-sectional area 0.02cm^2.Find the extension produced if length of wire is 5m.(Y=2×10^13 N/m^2​

Answer 1

Concept:

The Young's modulus of a material is defined as the ratio of tensile stress to tensile strain and indicates how easily it can stretch and flex.

Given:
Exerted force, $F=120 N$

cross- sectional area, $A=0.02cm^{2}=0.02*10^{-4}m^{2}$

length of wire, $L=5m$

Young's modulus for this metal, $Y=2*10^{13}N/m^{2}$

To Find:

We have to find the extension produced $\Delta L$

Solution:

By considering the given data,

Young's modulus is given by,

$Y=\frac{F/A}{\Delta L/L}$

$Y=\frac{F*L}{A*\Delta L}$

$\Delta L=\frac{120*6}{0.02*10^{-4}*2*10^{13} }=1.5*10^{-5} m$

Hence the extension produced if the length of wire is 5m is $1.5*10^{-5} m$

#SPJ2