What is the increase in area of the ring?

In summary, the increase in area of a metal ring with diameter D and coefficient of linear expansion y, when the temperature is increased by 1 degree Celsius, is [(pi)(D^2)y]/2. This can also be derived using the circumference and the square of the diameter. However, the given answer in the book is incorrect and should be 2(D^2)y instead.
  • #1
Amith2006
427
2
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.
 
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  • #2
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:

[tex]L = L_{0}(1 + \alpha\Delta T)[/tex] (1)

So,

[tex]L^{2} = L_{0}^2(1 + \alpha\Delta T)^2[/tex] (2)

For small values of [itex]\alpha\Delta T[/itex],

[tex]L^{2} \cong L_{0}^2(1 + 2\alpha\Delta T)[/tex] (3)

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

Clearly [itex]\beta = 2\alpha[/itex] so you could well have written it as

[tex]A = A_{0}(1 + \beta\Delta T)[/tex] (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

[itex]A_{0} = \pi D^{2}/4[/itex]

So,

[tex]\Delta A = A_{0}(2\alpha\Delta T)[/tex] (5)

[itex]\Delta T = 1[/itex], hence

[tex]\Delta A = \frac{\pi\alpha D^{2}}{2}[/tex] (6)

Lets work it out using circumference just to be sure

[tex]\pi D = \pi D_{0}(1 + \alpha\Delta T)[/tex] (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?
 
Last edited:

1. What is the increase in area of the ring?

The increase in area of the ring refers to the amount of additional space that is added to the original area of the ring. This is typically measured in square units, such as square inches or square centimeters.

2. How is the increase in area of the ring calculated?

The increase in area of the ring can be calculated by subtracting the original area from the final area. This can be done by using the formula A = πr², where A represents the area, and r represents the radius of the ring.

3. Can the increase in area of the ring be negative?

No, the increase in area of the ring cannot be negative. This is because an increase implies a positive change, and an area cannot have a negative value. If the final area is smaller than the original area, the increase in area would be equal to 0.

4. How does the increase in area of the ring affect the circumference?

The increase in area of the ring does not directly affect the circumference. The circumference of a ring is determined by its radius and is not affected by changes in its area. However, changes in the radius can indirectly affect the circumference.

5. Are there any real-life applications for understanding the increase in area of the ring?

Yes, understanding the increase in area of the ring can be useful in various fields such as engineering, architecture, and design. For example, in architecture, calculating the increase in area of a circular room can help determine the amount of material needed for flooring or painting.

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