The thermal stress in the rod is given by: \[ \text{Stress} = Y \alpha \Delta T, \] where:
\( Y = 2 \times 10^{11} \, \text{N/m}^2 \) (Young's modulus),
\( \alpha = 10^{-5} \, \text{K}^{-1} \) (coefficient of linear expansion),
\( \Delta T = 200 \, \text{K} \) (temperature change).
The force due to thermal stress is: \[ F = \text{Stress} \cdot A = Y \alpha \Delta T \cdot A. \] Substitute \( A = 10^{-4} \, \text{m}^2 \): \[ F = (2 \times 10^{11}) (10^{-5}) (200) (10^{-4}). \] Simplify: \[ F = 4 \times 10^4 \, \text{N}. \]
Final Answer: The compressive tension produced in the rod is: \[ \boxed{4 \times 10^4 \, \text{N}}. \]
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.