A rod, of length $$L$$ at room temperature and uniform area of cross section $$A$$, is made of a metal having coefficient of linear expansion $$\alpha$$ /$$^{\circ}$$C. It is observed that an external compressive force $$F$$ is applied on each of its ends, prevents any change in the length of the rod when its temperature rises by $$\Delta T$$ K. Young's modulus, $$Y$$ for this metal is:

We begin by recalling that when the temperature of a free rod rises by $$\Delta T$$, its natural tendency is to increase its length. The linear thermal expansion formula is first stated:

$$\Delta L_{\text{thermal}} = \alpha L \Delta T,$$

where $$\alpha$$ is the coefficient of linear expansion, $$L$$ is the original length and $$\Delta T$$ is the rise in temperature.

This change of length produces a thermal strain (fractional change in length) given by

$$\varepsilon_{\text{thermal}} = \frac{\Delta L_{\text{thermal}}}{L} = \alpha \Delta T.$$

In the problem, however, a compressive force $$F$$ is applied at both ends so that the rod is not allowed to change its length at all. Therefore, the total strain in the rod must be zero:

$$\varepsilon_{\text{total}} = \varepsilon_{\text{thermal}} + \varepsilon_{\text{mechanical}} = 0.$$

We have already written $$\varepsilon_{\text{thermal}} = \alpha \Delta T.$$ Next, we write the mechanical strain caused by the external compressive force. Young’s modulus $$Y$$ is defined by the relation

$$Y = \frac{\text{stress}}{\text{strain}} \quad \Longrightarrow \quad \text{strain} = \frac{\text{stress}}{Y}.$$

The compressive stress produced by the force $$F$$ over the cross-sectional area $$A$$ is

$$\sigma = \frac{F}{A}.$$

Because the force is compressive, the corresponding mechanical strain is also compressive (negative):

$$\varepsilon_{\text{mechanical}} = -\frac{\sigma}{Y} = -\frac{F}{A\,Y}.$$

Now we impose the condition that the rod’s length does not change, i.e. $$\varepsilon_{\text{total}} = 0$$. Substituting the expressions for the two strains, we get

$$\alpha \Delta T \;+\; \left(-\frac{F}{A\,Y}\right) = 0.$$

Simplifying,

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

We now solve for Young’s modulus $$Y$$ by cross-multiplying:

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

This expression matches Option C in the given list.

Hence, the correct answer is Option C.