Y=TLΔLAY=\frac{TL}{\Delta LA}Y=ΔLATL
As in both wire material is same hence the value of young's modulus will be same in both the cases
hence equating
200×31.2×10−3×A=200×6ΔL×A\frac{200\times3}{1.2\times10^{-3}\times A}=\frac{200\times6}{\Delta L\times A}1.2×10−3×A200×3=ΔL×A200×6
ΔL=2.4×10−3m=2.4mm\Delta L=2.4\times10^{-3}m=2.4mmΔL=2.4×10−3m=2.4mm
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