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: \[ \boxed{4 \times 10^4 \, \text{N}}. \]
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: