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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`
10 months 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)
```
10 months ago
```							ΔT = T2-T1ΔL/L=length =L=1%=1/100=0.01ΔL= αLΔTΔL/L= αxΔT0.01=αxΔTα=0.01/ΔTGiven Area of copper plate = 10Lx2L=20L^2By thermal expansion theory ΔA=βAΔTΔA/A=βΔTbut β=2 αΔA/A=2αΔT=2×0.01=2×10^-2ΔA/A×100=2×10^-2×100=2%
```
10 months ago
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