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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
anser 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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