Given here,
Length, L=1m and extension in length, ΔL=0.4×10−3m
Mass of load, m=1kg and diameter of wire, d=0.4×10−3m
Young's modulus is given by
Y=AΔLFL=(4πd2)0.4×10−3(mg)×(1)
⇒Y=π(0.4×10−3)2×0.4×10−310×4
⇒Y=22×64×10−3×10−940×7
⇒Y=0.199×10−12Nm−2
Error in measurement of Young's modulus,
YΔY=FΔF+LΔL+AΔA+(ΔL)Δ(ΔL)
=0.40.02+2dΔd=40.2+2×0.40.01
=20.1+20.1=0.1
=0.199×1011≈2×1010
Hence, the value of x=2.