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: \[ \boxed{4 \times 10^4 \, \text{N}}. \]
The term independent of $ x $ in the expansion of $$ \left( \frac{x + 1}{x^{3/2} + 1 - \sqrt{x}} \cdot \frac{x + 1}{x - \sqrt{x}} \right)^{10} $$ for $ x>1 $ is:

Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.
If $ \theta \in [-2\pi,\ 2\pi] $, then the number of solutions of $$ 2\sqrt{2} \cos^2\theta + (2 - \sqrt{6}) \cos\theta - \sqrt{3} = 0 $$ is: