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 $2 L \times L$ is heated from $T_{1}$ to $T_{2}$ ?

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Solution

Step 1: Given
Length 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 = $2 L \times L$
Assume the coefficient of linear expansion be $α$ .
Assume the coefficient of area (superficial) expansion be $β$ .

Step 2: Formula used and calculation
We know that
$\frac{∆ L}{L} = α ∆ T 0.01 = α (T_{2} - T_{1}) - - - - - (1)$

Area of copper plate = $2 L \times L$
On heating change in the area will be given by
$∆ A = β A ∆ T \frac{∆ A}{A} = β ∆ T$
We know that for a given material $β = 2 α$ .
Therefore,
$\frac{∆ A}{A} = β ∆ T \frac{∆ A}{A} = 2 α ∆ T \frac{∆ A}{A} = 2 α (T_{2} - T_{1}) \frac{∆ A}{A} = 2 \times 0.01 (F r o m e q (1)) \frac{∆ A}{A} = 0.02$
Therefore change in area will be $2 %$ .

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A thin copper wire of length L increases by 1% when heated from temperature T1 to T2. What is the percentage change in the area when a thin copper plate having dimensions 2L×L is heated from T1 to T2?

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 $2 L \times L$ is heated from $T_{1}$ to $T_{2}$ ?