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 /oC. 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 TK. Young's modulus, Y, for this metal is :

F \over A\alpha (\Delta T - 273)

Heat and Thermodynamics PYQ — JEE Main 2019 · 9 Jan · Shift 1 · Q49 with Solution

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Correct answer: A

If the rod were allowed to expand freely for a temperature rise of $\Delta T$ , then the change in length would be

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

So the thermal strain is

\epsilon_{\text{thermal}} = \frac{\Delta L_{\text{thermal}}}{L} = \alpha \Delta T

A compressive force $F$ is applied at each end. The internal compressive stress developed in the rod is

\sigma = \frac{F}{A}

The corresponding compressive strain is

\epsilon_{\text{compressive}} = \frac{\sigma}{Y} = \frac{F}{AY}

This strain acts opposite to thermal expansion.

The problem states that the rod does not change its length when temperature rises by $\Delta T$ . Hence,

\epsilon_{\text{thermal}} = \epsilon_{\text{compressive}}

So,

\alpha \Delta T = \frac{F}{AY}

Rearranging,

Y = \frac{F}{A\alpha \Delta T}

Thus the correct option is

\boxed{\frac{F}{A\alpha \Delta T}}

which is Option A.