Thermal Expansion - Practice Questions & MCQ

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Thermal Expansion and its types is considered one of the most asked concept.
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Coefficient of linear expansion of brass and steel rods are $\alpha_{1}$ and $\alpha_{2}$ . Lengths of brass and steel rods are $\l_{1}$ and $\l_{2}$ respectively. If ( $\l_{2}$ - $\l_{1}$ ) is maintained same at all temperatures, which one of the following relations holds good?

$\alpha_{1}\l_{2}=\alpha_{2}\l_{1}$

$\alpha_{1}\l_{2}^{2}=\alpha_{2}\l_{1}^{2}$

$\alpha_{1}^{2}\l_{2}=\alpha_{2}^{2}\l_{1}$

$\alpha_{1}\l_{1}=\alpha_{2}\l_{2}$

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Concepts Covered - 2

Thermal Expansion and its types

Thermal expansion is the tendency of material to change its shape, area, and volume in response to a change in temperature. So, if there is any change in temperature every material has tendency to change its dimension and the amount of change depends on the type of materials.

Thermal expansion is minimum in case of solids but maximum in case of gases because the intermolecular force is maximum in solids but minimum in gases.

So, solids can expand in one dimension, two dimension and three dimension while liquids and gases usually suffers change in volume only.

Thermal expanision is basically of three types -

Linear expansion : When a solid is heated and it's length increases, then the expansion is called linear expansion.

Let us take an specimen of length L₀. There is two scenario, first is before heating and the second image shows after heating. So,

$\\ {\text { (i) Change in the length of the specimen is } \Delta L=L_o \alpha \Delta T} \\ {\ \ \ \ \ \(Here,L=\text { Original length, } \Delta T=\text { Temperature change) }} \\ \\ {\text { (ii) Final length of the specimen is } L=L_o(1+\alpha \Delta T)} \\ \\ {\text { (iii) Co-efficient of linear expansion } \alpha=\frac{\Delta L}{L_{0} \Delta T}} \\ \\ {\text { (iv) Unit of } \alpha \text { is }^{\circ} \mathrm{C}^{-1} \text { or } K^{-1} \text { . It's dimension is }\left[\theta^{-1}\right]}$

Superficial (areal) expansion : When the temperature of a 2-Dimensiuonal specimen is changed, it's area changes, then the expansion is called superficial or areal expansion.

$\\ {\text { (i) Change in area is } \Delta A=A_o \beta \Delta T} \\ \\ {\ \ \ \ \ \ (A=\text { Original area, } \Delta T=\text { Temperature change) }} \\ \\ {\text { (ii) Final area } A=A_o(1+\beta \Delta T)} \\ \\ {\text { (iii) Co-efficient of superficial expansion } \beta=\frac{\Delta A}{A_{0} \Delta T}} \\ \\ {\text { (iv) Unit of } \beta \text { is }^{\circ} \mathrm{C} \text { or } K .}$

Volume or cubical expansion : When a 3-Dimensional solid is heated and it's volume increases, then the expansion is called volume or cubical expansion.

Now there is one relation between the $\alpha, \beta \ and\ \gamma$ , which can be written as -

$\alpha=\frac{\beta}{2}=\frac{\gamma}{3} \Rightarrow \alpha : \beta : \gamma=1 : 2 : 3$

Hence, for the same rise in temperature -

Percentage change in area = 2 times the percentage change in length.
Percentage change in volume = 3 times the percentage change in length.

Effects of thermal expansion on Solids

(1) Bi-metallic strip : When two strips of equal lengths but of different materials (such that they have different value of coefficient of linear expansion) when join together, it is called “bi-metallic strip”, and it can be used in thermostat to break or make electrical contact. Bi-metallic strip has the characteristic property of bending on heating. This is due to unequal linear expansion of the two metal. The strip will bend with metal of greater $\alpha$ on outer side.

The above figure shows the condition before and after heating the bi-metallic strip.

(2) Effect of temperature on the time period of a simple pendulum : Let us suppose a pendulum clock keeps proper time at temperature $\theta$ . If the temperature is increased to $\theta' \ (\theta'>\theta)$ then due to linear expansion, length of pendulum and from the formula, we knows that the time period of simple pendulum is directly proportional to the square root of length of the pendulum hence its time period will increase.
Fractional change in time period $\frac{\Delta T}{T}=\frac{1}{2} \alpha \Delta \theta$

(i) In summer, the temperature will rise and due to this there will be increment in its time period. A pendulum clock becomes
slow and will lose time.
Loss of time in a time period is given by - $\Delta T=\frac{1}{2} \alpha \Delta \theta T$
(ii) Time lost by the clock in a day - $\Delta T=\frac{1}{2} \alpha \Delta \theta t=\frac{1}{2} \alpha \Delta \theta(86400)=43200 \alpha \Delta \theta \sec$