A steel rail of length 5 m and area of cross section 40 cm$$^2$$ is prevented from expanding along its length while the temperature rises by 10°C. If coefficient of linear expansion and Young's modulus of steel are $$1.2 \times 10^{-5}$$ K$$^{-1}$$ and $$2 \times 10^{11}$$ N m$$^{-2}$$ respectively, the force developed in the rail is approximately:

We know that any material, if free to expand, would increase its length by the thermal relation

$$\Delta L = \alpha L \Delta T,$$

where $$\alpha$$ is the coefficient of linear expansion, $$L$$ is the original length and $$\Delta T$$ is the rise in temperature. Here, however, the rail is rigidly fixed, so no actual change in length is allowed. This restriction produces a compressive mechanical strain exactly equal in magnitude (but opposite in sense) to the free thermal strain.

So the strain locked inside the rail is

$$\text{strain} = \alpha \Delta T.$$

Young’s modulus $$Y$$ links stress and strain through the formula

$$\text{stress} = Y \times \text{strain}.$$

Substituting the expression for strain, we obtain

$$\text{stress} = Y \, \alpha \, \Delta T.$$

The numerical values given are

$$Y = 2 \times 10^{11}\ \text{N m}^{-2}, \qquad \alpha = 1.2 \times 10^{-5}\ \text{K}^{-1}, \qquad \Delta T = 10^\circ\text{C}.$$

Putting these into the stress formula, we have

$$\text{stress} = \left(2 \times 10^{11}\right)\left(1.2 \times 10^{-5}\right)(10) = 2 \times 1.2 \times 10^{11} \times 10^{-5} \times 10.$$

Now, $$1.2 \times 10^{-5} \times 10 = 1.2 \times 10^{-4},$$ so

$$\text{stress} = 2 \times 1.2 \times 10^{11} \times 10^{-4} = 2.4 \times 10^{7}\ \text{N m}^{-2}.$$

Force is obtained from stress through the relation

$$\text{force} = \text{stress} \times \text{area}.$$

The cross-sectional area is 40 cm$$^2$$. Converting this to square metres:

$$40\ \text{cm}^2 = 40 \times 10^{-4}\ \text{m}^2 = 4.0 \times 10^{-3}\ \text{m}^2.$$

Substituting the values,

$$\text{force} = (2.4 \times 10^{7}) \times (4.0 \times 10^{-3}) = 2.4 \times 4.0 \times 10^{7} \times 10^{-3}.$$

Since $$10^{7} \times 10^{-3} = 10^{4},$$ we get

$$\text{force} = 9.6 \times 10^{4}\ \text{N}.$$

Rounding to one significant figure, this is

$$\text{force} \approx 1 \times 10^{5}\ \text{N}.$$

Hence, the correct answer is Option D.