The applied load =F=mg=2×9.8=19.6N
diameter =1mm=1×10
−3
m
=> radius=0.5×10
−7
m
Area , A=3.14r
2
=3.14×0.25×10
−6
=0.785×10
−6
We know that,
stress=Force/area
=19.6/0.785×10
−6
24.96×10
−6
change in length,ΔL=1mm=10
−3
m
Length,L=3m
hencestrain=ΔL/L
=10
−3
/3
we know that young's modulus is given by the ratio of stress and strain,
Hence,
E=24.96×10
−6
/10
−3
/3
=74.8×10
9
=7.48×10
10
Nm
−2
