A rod of iron of Young’s modulus Y= 2.0 x10¹¹ N / m² just fits the gap between two rigid supports 1m apart. If the rod is heated through 100°C the strain energy of the rod is (α =18 x 10⁻⁶ °C⁻¹ and area of cross-section A = 1 cm² )

The correct option is 32.4 J.

Step-by-step Explanation:

When a rod is heated, it tends to expand. However, if the rod is fixed between two rigid supports, it is prevented from expanding, which generates thermal stress and strain within the rod. This stored energy is called strain energy.

Let's first list the given values:
Young's modulus of iron, $Y=2.0\times {10}^{11}{\text{N/m}}^2$
Length of the rod, $L=1\text{m}$
Increase in temperature, $\Delta T=100\text{°C}$
Coefficient of linear expansion, $\alpha =18\times {10}^{-6}{\text{°C}}^{-1}$
Area of cross-section, $A=1{\text{cm}}^2=1\times {10}^{-4}{\text{m}}^2$

1. Calculate the Volume of the Rod:
The volume $V$ of the rod is the product of its cross-sectional area and its length:
$V=A\times L={10}^{-4}{\text{m}}^2\times 1\text{m}={10}^{-4}{\text{m}}^3$

2. Calculate the Thermal Strain:
Thermal strain is given by the formula:
$\text{Strain}=\alpha \Delta T$
Substituting the given values:
$\text{Strain}=18\times {10}^{-6}\times 100=18\times {10}^{-4}$

3. Calculate the Strain Energy:
The formula for the strain energy $U$ stored in the rod is:
$U=\frac{1}{2}\times Y\times {\left(\text{Strain}\right)}^2\times V$
Substituting the values we obtained:
$U=\frac{1}{2}\times (2.0\times {10}^{11})\times {(18\times {10}^{-4})}^2\times {10}^{-4}$
Simplify the equation:
$U={10}^{11}\times (324\times {10}^{-8})\times {10}^{-4}$
Combine the powers of 10:
$U=324\times {10}^{11-8-4}$
$U=324\times {10}^{-1}=32.4\text{J}$

Thus, the strain energy of the rod is 32.4 J.