Question

# A thin copper wire of length $L$ increases by $1%$ when heated from temperature ${T}_{1}$ to ${T}_{2}$. What is the percentage change in the area when a thin copper plate having dimensions $2L×L$ is heated from ${T}_{1}$ to ${T}_{2}$?

Open in App
Solution

## Step 1: GivenLength of the copper wire = $L$Percentage change in the length, $\frac{∆L}{L}=1%=\frac{1}{100}=0.01$Initial temperature = ${T}_{1}$Final temperature = ${T}_{2}$Dimension of copper plate = $2L×L$Assume the coefficient of linear expansion be $\alpha$.Assume the coefficient of area (superficial) expansion be $\beta$. Step 2: Formula used and calculationWe know that$\frac{∆L}{L}=\alpha ∆T\phantom{\rule{0ex}{0ex}}0.01=\alpha \left({T}_{2}-{T}_{1}\right)-----\left(1\right)$ Area of copper plate = $2L×L$On heating change in the area will be given by$∆A=\beta A∆T\phantom{\rule{0ex}{0ex}}\frac{∆A}{A}=\beta ∆T$We know that for a given material $\beta =2\alpha$.Therefore, $\frac{∆A}{A}=\beta ∆T\phantom{\rule{0ex}{0ex}}\frac{∆A}{A}=2\alpha ∆T\phantom{\rule{0ex}{0ex}}\frac{∆A}{A}=2\alpha \left({T}_{2}-{T}_{1}\right)\phantom{\rule{0ex}{0ex}}\frac{∆A}{A}=2×0.01\left(Fromeq\left(1\right)\right)\phantom{\rule{0ex}{0ex}}\frac{∆A}{A}=0.02\phantom{\rule{0ex}{0ex}}$Therefore change in area will be $2%$.

Suggest Corrections
0
Explore more