A compressive force, $$F$$ is applied at the two ends of a long thin steel rod. It is heated, 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:

We are told that a long, thin steel rod of initial length $$l$$ and cross-sectional area $$A$$ is subjected simultaneously to two different actions. One action is purely mechanical: a compressive force $$F$$ is applied axially at its ends. The other action is purely thermal: the rod is heated so that its temperature rises by $$\Delta T$$. Experimentally, we observe that the combined effect of these two actions leaves the overall length of the rod unchanged. Mathematically, this means that the total, or net, linear strain is zero.

We first recall the definition of linear strain. Whenever a body undergoes a fractional change in length, that fraction is called the strain: $$\text{strain} = \dfrac{\Delta l}{l}.$$ If two kinds of strain act simultaneously, we simply add them (with the proper algebraic signs) to get the net strain.

We therefore identify the two individual strains acting on the rod:

1. Thermal strain due to heating. The coefficient of linear expansion of the material is $$\alpha$$. By definition, the fractional change in length produced solely by a temperature rise $$\Delta T$$ is

$$\text{thermal strain} \;=\; \alpha \,\Delta T.$$

This strain is positive because heating tends to elongate the rod.

2. Mechanical strain due to the compressive force. We start with the relation between stress, strain and Young’s modulus. Young’s modulus $$Y$$ of a material is defined by the formula

$$Y \;=\; \dfrac{\text{stress}}{\text{strain}}.$$

Here the axial stress caused by the applied force $$F$$ is simply the force divided by the area of cross-section:

$$\text{stress} \;=\; \dfrac{F}{A}.$$

Substituting this stress into the defining equation of Young’s modulus, we get the mechanical strain:

$$\text{strain}_{\text{mech}} \;=\; \dfrac{\text{stress}}{Y} \;=\; \dfrac{F/A}{Y} \;=\; \dfrac{F}{A\,Y}.$$

Because the applied force is compressive, it attempts to shorten the rod. Hence this mechanical strain is negative with respect to the original length. To indicate the sign explicitly, we may write it as $$-\dfrac{F}{A\,Y}.$$

Adding the two strains and imposing the zero-net-strain condition. The net strain is the algebraic sum of the thermal strain and the mechanical strain. Setting their sum to zero gives

$$\text{thermal strain} \;+\; \text{mechanical strain} \;=\; 0,$$

$$\alpha \,\Delta T \;-\; \dfrac{F}{A\,Y} \;=\; 0.$$

Now we solve this equation step by step for the unknown force $$F$$. First, move the mechanical-strain term to the right-hand side:

$$\alpha \,\Delta T \;=\; \dfrac{F}{A\,Y}.$$

Next, multiply both sides by $$A\,Y$$ to isolate $$F$$:

$$F \;=\; A\,Y \, \alpha \,\Delta T.$$

Thus the compressive force necessary to counteract the expansion produced by the temperature rise and leave the rod’s length unchanged is

$$F \;=\; A Y \alpha \Delta T.$$

We compare this result with the given options. It matches Option B.

Hence, the correct answer is Option B.