Related Formula
T = (2 m₁ m₂)/(m₁ + m₂) g$$T = \frac{2 m_1 m_2}{m_1 + m_2} g$$
Strain = (Δ )/( ) = (F)/(A Y) = (T)/(A 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 = ( (2 × 2 × 4)/(2 + 4) ) g = ( (16)/(6) ) × 10 = (160)/(6) = (80)/(3) 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 × 10⁻⁵ m$r = 4.0 \times 10^{-5} \text{ m}$.
A = π r² = π (4 × 10⁻⁵)² = 16π × 10⁻¹⁰ m²$$A = \pi r^2 = \pi (4 \times 10^{-5})^2 = 16\pi \times 10^{-10} \text{ m}^2$$
Step 3: Calculate Strain
Strain = (T)/(A Y) = 80/3(16π × 10⁻¹⁰) × (2 × 10¹¹)$$\text{Strain} = \frac{T}{A Y} = \frac{80/3}{(16\pi \times 10^{-10}) \times (2 \times 10^{11})}$$
Strain = (80/3)/(32π × 10) = (80)/(3 × 320π) = (1)/(12π)$$\text{Strain} = \frac{80/3}{32\pi \times 10} = \frac{80}{3 \times 320\pi} = \frac{1}{12\pi}$$
Step 4: Extract Alpha
Comparing with (1)/(α π)$\frac{1}{\alpha \pi}$:
α = 12$\alpha = 12$
Pattern Recognition
Linkage problem: Mechanics (Atwood tension) →$\to$ Solids (Strain). Memorize Atwood tension T = 2m₁m₂g/(m₁+m₂)$T = 2m_1m_2g/(m_1+m_2)$ to skip drawing free body diagrams during exams.
Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Class 11 Physics: Laws of Motion