Given, L0=20mL_0=20mL0=20m ,Ti=100C,Tf=500CT_i=100C,T_f=500CTi=100C,Tf=500C,thus ΔT=Tf−Ti=400K\Delta T=T_f-T_i=400KΔT=Tf−Ti=400K.
ΔL=L−L0,&αL=1.2×10−5K−1\Delta L=L-L_0,\& \alpha_L=1.2 \times 10^{-5}K^{-1}ΔL=L−L0,&αL=1.2×10−5K−1 ,
We know that
Thus,
Therefore, expansion of length is ΔL=0.096m\Delta L=0.096mΔL=0.096m
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