A rod of length 1000 mm and co-efficient of linear expansion $\mathrm{\alpha } = {10}^{-4}$ per degree is placed symmetrically between fixed walls separated by l00l mm. The Young's modulus of the rod is ${10}^{11}\mathrm{N}/{\mathrm{m}}^2$. If the temperature is increased by 20°C, then the stress developed in the rod is (in $\mathrm{N}/{\mathrm{m}}^2$)

The change in length of rod due to increase in temperature in absence of walls is

$∆\mathrm{l} = \mathrm{l\alpha } ∆\mathrm{T} = 1000\times {10}^{-4}\times 20 \mathrm{mm}$

      = 2 mm

But the rod can expand upto 1001 mm only.
At that temperature its natural length is = 1002 mm.
$\therefore$ compression = l mm 
$\therefore$  mechanical stress

$= \mathrm{Y} = \frac{∆\mathrm{l}}{\mathrm{l}} = {10}^{11}\times \frac{1}{1000} = {10}^8 \mathrm{N}/{\mathrm{m}}^2$