A thin copper wire of length l increase in lenght by 1% when heated from T¹ to T² what is the percentage change in area when a thin copper plate having dimension (10L×2L) is heated from T¹ to T²
Mansi , 5 Years ago
Grade 11
2 Answers
Arun
Last Activity: 5 Years ago
See, suppose we have l* as the initial length and l being the final length Hence, l=l* x (1+αΔt); where Δt= t2-t1 and α=coefficient of linear expansion or, l=l* + l* x αΔt or, (l-l*)/l*=αΔt=1/100 So, A=A* x (1+βΔt); //b=coefficient of area expansion or (A-A*)/A* =βΔt =2αΔt=2/100=2% (β=2α)
Khimraj
Last Activity: 5 Years ago
ΔT = T2-T1 ΔL/L=length =L=1%=1/100=0.01
ΔL= αLΔT ΔL/L= αxΔT 0.01=αxΔT α=0.01/ΔT
Given Area of copper plate = 10Lx2L=20L^2 By thermal expansion theory ΔA=βAΔT ΔA/A=βΔT but β=2 α ΔA/A=2αΔT =2×0.01 =2×10^-2 ΔA/A×100=2×10^-2×100=2%
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