Given parameters:
L=3 m
ΔL=3×10−3 m
V=600×10−6 m3
Y=1.1×1011 N/m2
Strain is given by:
Strain=LΔL=33×10−3=10−3
Elastic potential energy stored in the wire is:
U=21×Stress×Strain×Volume
Since Stress=Y×Strain:
U=21×Y×(Strain)2×V
Substituting the values:
U=21×(1.1×1011)×(10−3)2×(600×10−6)
U=21×1.1×1011×10−6×600×10−6
U=21×1.1×600×10−1
U=21×66=33 J
Answer: 33