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: \[ \boxed{4 \times 10^4 \, \text{N}}. \]
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.