You are given a square plate of side $1 m$ . The coefficient of linear expansion $α$ , obtained the relation between the coefficient of superficial expansion $β$ and linear expansion $α$ for the unit rise in temperature.

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Solution

Step 1: We have
The initial side length of square $= L_{i} = 1 m$
Coefficient of linear expansion $= α$
Temperature rise $= ∆ T = 1 K$
Step 2: To find the relation between the coefficient of superficial expansion $β$ and linear expansion $α$ for the unit rise in temperature.
The initial area of square $= A_{i} = {(L_{i})}^{2}$
$A_{i} = 1 m^{2}$
Step 3: To find new length and new area of square
On heating,
The new length of square $= L_{f} = L_{i} (1 + α ∆ T)$
$L_{f} = L_{i} + L_{i} α ∆ T$
$L_{f} - L_{i} = L_{i} α ∆ T$
$∆ L = L_{i} α ∆ T$
So, $α = \frac{∆ L}{L_{i} ∆ T} = \frac{∆ L}{∆ T} . . . . . . . . . . (1) [L_{i} = 1 m]$
The new area of the square $= A_{f} = {(L_{f})}^{2}$
$A_{f} = {(L_{i} (1 + α ∆ T))}^{2}$
$A_{f} = {L_{i}}^{2} {(1 + α ∆ T)}^{2}$
$A_{f} = A_{f} (1 + 2 α ∆ T) [S i n e, {(1 + x)}^{n} = (1 + n x)]$
$A_{f} - A_{i} = 2 A_{i} α ∆ T$
$\frac{∆ A}{A_{i} ∆ T} = 2 α . . . . . . . . . . (2)$
We know that
$\frac{∆ A}{A_{i} ∆ T} = β . . . . . . . . . (3)$
From equation $(2) a n d (3)$
We get
$β = 2 α$
Hence, the required relationship between the coefficient of superficial expansion $β$ and linear expansion $α$ for the unit rise in temperature is $β = 2 α$ .

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You are given a square plate of side 1m. The coefficient of linear expansion α, obtained the relation between the coefficient of superficial expansion β and linear expansion α for the unit rise in temperature.

You are given a square plate of side $1 m$ . The coefficient of linear expansion $α$ , obtained the relation between the coefficient of superficial expansion $β$ and linear expansion $α$ for the unit rise in temperature.