What is the increase in area of the ring?

[Amith2006]

Sir,
The diameter of a metal ring is D and the coefficient of linear expansion is y. The temperature of the ring is increased by 1 degree Celsius. What is the increase in area of the ring?
I solved it the following way:
Coefficient of superficial expansion = 2y
Initial area of the ring(A) = [(pi)D^2]/4
Increase in diameter of the ring = Dy
Increase in area of the ring = A x (2y)
= [(pi)D^2]/4 x 2y
= [(pi)(D^2)y]/2
But the answer given in my book is 2(D^2)y. Please help. Here the symbol ^ stands for power.

 

[maverick280857]

The fundamental "metric" to consider for area expansion should be length rather than area itself. We can show that for small temperature changes (specifically when the product of the expansion coefficient and the temperature change is much smaller than unity) the results obtained from either approach are equivalent:

$$L = L_{0}(1 + \alpha\Delta T)$$ (1)

So,

$$L^{2} = L_{0}^2(1 + \alpha\Delta T)^2$$ (2)

For small values of $\alpha\Delta T$,

$$L^{2} \cong L_{0}^2(1 + 2\alpha\Delta T)$$ (3)

(I've used the Binomial Theorem here in the form $(1+x)^{n} \cong 1 + nx$ for $x << 1$)

Clearly $\beta = 2\alpha$ so you could well have written it as

$$A = A_{0}(1 + \beta\Delta T)$$ (4)

directly, but what we're saying is that this follows from length expansion. But we are working in the realm of small temperature changes anyway so that our model is linear (and the coefficients of expansion are all independent of temperature).

Now, let's come to your problem:

Let

$A_{0} = \pi D^{2}/4$

So,

$$\Delta A = A_{0}(2\alpha\Delta T)$$ (5)

$\Delta T = 1$, hence

$$\Delta A = \frac{\pi\alpha D^{2}}{2}$$ (6)

Lets work it out using circumference just to be sure

$$\pi D = \pi D_{0}(1 + \alpha\Delta T)$$ (7)

Area is proportional to the square of D, so squaring both sides, we get back equation (4) (using the Binomial Theorem). So the answer we're getting is correct. Perhaps a misprint. Which book is this?