A rod is fixed between two points at 20°C . The coefficient of linear expansion of material of rod is 1.1 x 10⁻⁵ / °C and Young’s modulus is 1.2 x10¹¹ N / m² . Find the stress developed in the rod if temperature of rod becomes 10°C

To find the thermal stress developed in the rod when its temperature decreases, we can follow a step-by-step physical derivation.

1. Identify the given parameters:
Initial temperature, $T_1=20\mathrm{°C}$
Final temperature, $T_2=10\mathrm{°C}$
Change in temperature,
$\mathrm{\Delta T}=T_1-T_2=20\mathrm{°C}-10\mathrm{°C}=10\mathrm{°C}$
Coefficient of linear expansion, $\alpha =1.1\times {10}^{-5}/\mathrm{°C}$
Young's modulus of the material, $Y=1.2\times {10}^{11}{\mathrm{N/m}}^2$

2. Understand the concept of Thermal Strain:
When a rod of original length $L$ undergoes a temperature change $\mathrm{\Delta T}$, its change in length $\mathrm{\Delta L}$ if it were free to expand or contract is given by:
$\mathrm{\Delta L}=L\alpha \mathrm{\Delta T}$
Since the rod is fixed firmly between two rigid points, it is prevented from contracting as the temperature drops. This prevention of contraction introduces a thermal strain in the rod, which is defined as:
$\text{Strain}=\frac{\mathrm{\Delta L}}{L}=\alpha \mathrm{\Delta T}$

3. Calculate the Thermal Stress:
According to Hooke's Law, Young's modulus ($Y$) is the ratio of stress to strain:
$Y=\frac{\text{Stress}}{\text{Strain}}$
Therefore, the thermal stress developed in the rod can be expressed as:
$\text{Stress}=Y\times \text{Strain}=Y\alpha \mathrm{\Delta T}$

4. Substitute the values:
Let's plug in the given values into the formula:
$\text{Stress}=(1.2\times {10}^{11})\times (1.1\times {10}^{-5})\times 10$
Simplifying the calculation step-by-step:
$\text{Stress}=1.2\times 1.1\times 10\times {10}^{11-5}$
$\text{Stress}=1.32\times 10\times {10}^6$
$\text{Stress}=1.32\times {10}^7{\text{N/m}}^2$

Thus, the thermal stress developed in the rod is $1.32\times {10}^7{\text{N/m}}^2$, which matches the option 1.32 x 10⁷ N / m².