Question

A load of 5 kg is attached to the free end of a wire of length 2 m and diameter 0.6 mm. If the extension of the wire is 0.2 mm, calculate the Young's modulus of the material of the wire. (g = 9.8 ms-2)​

Answer 1

Answer:

   E=1.73×$10^{-6}$N$m^{-2}$

Explanation:

Given Data:

mass=m=5kg

length=L=2m

Extensions=ΔL=0.2mm=2 ×$10^{2}$m

diameter=0.6mm=6×$10^{2}$m

radius=r=d/2=6×$10^{2}$m/2=3×$10^{2}$m

g=9.8ms-2  

Solution:

we know that young's modulus is given by the ratio of stress and strain,

Hence,

E=stress/strain

stress=F/A

F=mg=5 ×9.8=49 N

A=π$r^{2}$=3.14 ×3×$10^{2}$m× 3×$10^{2}$m=28.26×$10^{4}$$m^{2}$

stress=49/28.26×$10^{4}$$m^{2}$=1.73×$10^{-4}$N$m^{-2}$

Strain=ΔL/L =2 ×$10^{2}$m/2=1 ×$10^{2}$m

E=stress/Strain=1.73×$10^{-4}$/1 ×$10^{2}$m

                               E=1.73×$10^{-6}$N$m^{-2}$