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Worked Examples · Example 10.1

Q.Show that the coefficient of area expansion, (ΔA/A)/ΔT(\Delta A/A)/\Delta T, of a rectangular sheet of the solid is twice its linear expansivity, αl\alpha_l.

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For a rectangular sheet, area expansion comes from both length and width expanding independently. Since each linear dimension expands by a factor (1+αlΔT)(1 + \alpha_l \Delta T), the area expands by (1+αlΔT)2≈1+2αlΔT(1 + \alpha_l \Delta T)^2 \approx 1 + 2\alpha_l \Delta T, so the area expansion coefficient αA=2αl\alpha_A = 2\alpha_l.

Figure 10.8
Figure 10.8

The key insight is that area expansion isn't a separate phenomenon — it's just linear expansion happening in two perpendicular directions at once. When a solid is heated uniformly, every linear dimension expands according to the same coefficient αl\alpha_l. For a rectangle, that means both its length and its width increase, and the new area is simply the product of the two expanded dimensions.

Let’s walk through this carefully.

  1. Start with the definition of linear expansivity. The linear expansion coefficient αl\alpha_l tells you the fractional change in length per degree temperature change:

αl=1LdLdT\alpha_l = \frac{1}{L} \frac{dL}{dT}

For a finite temperature change ΔT\Delta T, if αl\alpha_l is constant (which it is, to a very good approximation for small ΔT\Delta T), the new length is:

L′=L(1+αlΔT)L' = L (1 + \alpha_l \Delta T)

This is the fundamental relation we’ll use.

  1. Now consider a rectangular sheet. Let the original length be aa and original breadth be bb. The original area is:

A=abA = a b

  1. After heating by ΔT\Delta T, both dimensions expand. The new length and breadth become:

a′=a(1+αlΔT)a' = a (1 + \alpha_l \Delta T)

b′=b(1+αlΔT)b' = b (1 + \alpha_l \Delta T)

So the new area is:

A′=a′b′=ab(1+αlΔT)2A' = a' b' = a b (1 + \alpha_l \Delta T)^2

  1. Expand the square and simplify.

A′=A(1+2αlΔT+αl2(ΔT)2)A' = A \left(1 + 2\alpha_l \Delta T + \alpha_l^2 (\Delta T)^2\right)

For typical solids, αl\alpha_l is of the order 10−5 K−110^{-5} \, \text{K}^{-1}, so αl2(ΔT)2\alpha_l^2 (\Delta T)^2 is utterly negligible compared to 2αlΔT2\alpha_l \Delta T for any reasonable ΔT\Delta T (say, up to a few hundred degrees). We drop that term:

A′≈A(1+2αlΔT)A' \approx A \left(1 + 2\alpha_l \Delta T\right)

  1. Read off the area expansion coefficient. The change in area is: …

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