### Related Formula T = frac2 m_1 m_2m_1 + m_2 g$T = \frac{2 m_1 m_2}{m_1 + m_2} g$ textStrain = fracDelta ellell = fracFA Y = fracTA Y$\text{Strain} = \frac{\Delta \ell}{\ell} = \frac{F}{A Y} = \frac{T}{A Y}$ ### Core Logic First, determine the tension T$T$ in the wire produced by the accelerating system (Atwood machine). Then, apply the formula for longitudinal strain using the calculated tension, cross-sectional area, and Young's Modulus. ### Step 1: Calculate Tension T = left( frac2 times 2 times 42 + 4 right) g = left( frac166 right) times 10 = frac1606 = frac803 text N$T = \left( \frac{2 \times 2 \times 4}{2 + 4} \right) g = \left( \frac{16}{6} \right) \times 10 = \frac{160}{6} = \frac{80}{3} \text{ N}$ ### Step 2: Calculate Area Given radius r = 4.0 times 10^-5 text m$r = 4.0 \times 10^{-5} \text{ m}$. A = pi r^2 = pi (4 times 10^-5)^2 = 16pi times 10^-10 text m^2$A = \pi r^2 = \pi (4 \times 10^{-5})^2 = 16\pi \times 10^{-10} \text{ m}^2$ ### Step 3: Calculate Strain textStrain = fracTA Y = frac80/3(16pi times 10^-10) times (2 times 10^11)$\text{Strain} = \frac{T}{A Y} = \frac{80/3}{(16\pi \times 10^{-10}) \times (2 \times 10^{11})}$ textStrain = frac80/332pi times 10 = frac803 times 320pi = frac112pi$\text{Strain} = \frac{80/3}{32\pi \times 10} = \frac{80}{3 \times 320\pi} = \frac{1}{12\pi}$ ### Step 4: Extract Alpha Comparing with frac1alpha pi$\frac{1}{\alpha \pi}$: alpha = 12$\alpha = 12$ ### Pattern Recognition Linkage problem: Mechanics (Atwood tension) to$\to$ Solids (Strain). Memorize Atwood tension T = 2m_1m_2g/(m_1+m_2)$T = 2m_1m_2g/(m_1+m_2)$ to skip drawing free body diagrams during exams. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids Class 11 Physics: Laws of Motion