পদার্থ বিজ্ঞান ১ম পত্র প্রশ্ন

AI-তৈরি নমুনা উত্তর

Given:
Side length of the cube, $L = 0.1\text{ m}$
Cross-sectional area of a face, $A = L^2 = (0.1\text{ m})^2 = 10^{-2}\text{ m}^2$
Initial volume, $V = L^3 = (0.1\text{ m})^3 = 10^{-3}\text{ m}^3$
Applied normal force, $F = 89.67 \times 10^5\text{ N}$
Displacement, $\Delta L = 3.05 \times 10^{-3}\text{ m}$

1. Bulk Strain:
Assuming lateral expansion is constrained or considering one-dimensional volume change, the change in volume is:
$\Delta V = A \cdot \Delta L$
$\text{Bulk strain} = \frac{\Delta V}{V} = \frac{\Delta L}{L} = \frac{3.05 \times 10^{-3}\text{ m}}{0.1\text{ m}} = 0.0305 = 3.05 \times 10^{-2}$

2. Bulk Modulus ($K$):
$\text{Normal stress} = \frac{F}{A} = \frac{89.67 \times 10^5\text{ N}}{10^{-2}\text{ m}^2} = 8.967 \times 10^8\text{ N/m}^2$
$K = \frac{\text{Stress}}{\text{Bulk strain}} = \frac{8.967 \times 10^8\text{ N/m}^2}{0.0305} \approx 2.94 \times 10^{10}\text{ N/m}^2$

3. Stiffness Coefficient ($k$):
$k = \frac{F}{\Delta L} = \frac{89.67 \times 10^5\text{ N}}{3.05 \times 10^{-3}\text{ m}} \approx 2.94 \times 10^9\text{ N/m}$