Question

A compressive force, $F$ is applied at the two ends of a long thin steel rod. It is hated simultaneously, such that its temperature increases by $\Delta T$. The net change in its length is zero. Let $l$ be the length of the rod, $A$ its area of cross-section, $Y$ its Young’s modulus, and $\alpha $ its coefficient of linear expansion. Then, $F$ is equal to:
(A) ${l^2}Y\alpha \Delta T$
(B) ${l^2}AY\alpha \Delta T$
(C) $\dfrac{{AY}}{{\alpha \Delta T}}$
(D) $AY\alpha \Delta T$

Answer 1

Hint: This question deals with the thermal expansion, particularly linear thermal expansion which deals with the change in length i.e. expansion of length. The young’s modulus and the particulars of linear expansion of a rod is given. To find the force we should relate the young’s modulus with the linear thermal expansion of the rod.

Complete step by step answer
Thermal expansion is a naturally occurring phenomenon. It occurs when an object expands due to a change in the temperature of the object. Temperature is the movement of the energy of the molecules in a substance. If an object has higher temperature then that the molecules are moving faster on the average. If we heat up an object, the molecules move faster and hence, they end up occupying more space. As a result of this the size of the object increases.
Thermal expansion is the tendency of an object to change in its area, volume as well as shape in response to change in temperature. Heating up a substance increases the kinetic energy of the substance.
Thermal expansion paves way for changes in dimension either in length or volume or area of the substance. Therefore, there are three types of thermal expansions namely Linear expansion, Area expansion, and Volume expansion.
Here in the problem we are provided with linear expansion
Linear expansion occurs when there is a change in length.
The linear expansion is given by the formula,
$ \Rightarrow \dfrac{{\Delta L}}{{{L_0}}} = \alpha \Delta T$
Where,
$\Delta L$ is the change in length
${L_0}$ is the original length of the object
$\alpha $ is the length expansion coefficient
$\Delta T$ is the change in temperature
The young’s modulus is the ratio of longitudinal stress by longitudinal strain
The formula for calculating the young’s modulus is given by
$Y = \dfrac{{F{L_0}}}{{A\Delta L}}$
Y is the young’s modulus
F is the force exerted by the object
$\Delta L$ is the change in length
${L_0}$ is the original length of the object
A is the area of the object
Given,
A compressive force, $F$ is applied at the two ends of a long thin steel rod. It is hated simultaneously, such that its temperature increases by $\Delta T$. The net change in its length is zero. Let $l$ be the length of the rod, $A$ its area of cross-section, $Y$ its Young’s modulus, and $\alpha $ its coefficient of linear expansion.
The force applied to the steel rod is $F$
The change in temperature is $\Delta T$
The length of the rod is $l$
The change in length $\Delta L = 0$
Area of the rod is $A$
The young’s modulus is Y
The coefficient of linear expansion is $\alpha $
To find the force,
First,
The linear expansion is given by the formula,
$ \Rightarrow \dfrac{{\Delta L}}{{{L_0}}} = \alpha \Delta T{\text{ }} \to {\text{1}}$
Secondly,
The young’s modulus is given by
$ \Rightarrow Y = \dfrac{{F{L_0}}}{{A\Delta L}}$
$ \Rightarrow \Delta LY = \dfrac{{F{L_0}}}{A}$
$ \Rightarrow \dfrac{{\Delta L}}{{{L_0}}} = \dfrac{F}{{AY}}{\text{ }} \to {\text{2}}$
The LHS of 1 and 2 are equal
Equating the RHS of 1 and 2 we get,
$ \Rightarrow \dfrac{F}{{AY}} = \alpha \Delta T$
$ \Rightarrow F = AY\alpha \Delta T$

Hence the correct answer is option (D) $AY\alpha \Delta T$

Note Here in the question it is given that the change in net length is zero. Most of the students get confused and put zero for the value of change in length, if you do that you cannot equate the two equations and you won’t get an answer.