Two blocks of mass 2 kg and 4 kg are connected by a metal wire going over a smooth pulley as shown in figure. The radius of wire is 4.0 × 10⁻⁵ m and Young's modulus of the metal is 2.0 × 10¹¹ N/m². The longitudinal strain developed in the wire is (1)/(απ). The value of α is _________. [Use g = 10 m/s² ] Numerical answer type. Work it out, then open the answer.

Answer: 12. Atwood machine tension: T = (2m₁ m₂ g)/(m₁ + m₂) = (2 × 2 × 4 × 10)/(6) = 80/3 N A = π r² = π (4 × 10⁻⁵)² = 16π × 10⁻¹⁰ m² Strain = T/AY = 80/316π × 10⁻¹⁰ × 2 × 10¹¹ = (80/3)/(320π) = (1)/(12π) α = 12

Two blocks of mass 2 kg and 4 kg are connected by a metal wire going… | JEE Main 2024 (31 Jan, Shift 2)

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Elastic Behaviour, Stress & Strain

Two blocks of mass $2\text{ kg}$ and $4\text{ kg}$ are connected by a metal wire going over a smooth pulley as shown in figure. The radius of wire is $4.0 \times 10^{-5}\text{ m}$ and Young's modulus of the metal is $2.0 \times 10^{11}\text{ N/m}^2$ . The longitudinal strain developed in the wire is $\frac{1}{\alpha\pi}$ . The value of $\alpha$ is _________. [Use $g = 10\text{ m/s}^2$ ]

Numerical answer type. Work it out, then open the answer.

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Answer: 12

Atwood machine tension: $T = \frac{2m_1 m_2 g}{m_1 + m_2} = \frac{2 \times 2 \times 4 \times 10}{6} = \frac{80}{3}\ \text{N}$
$A = \pi r^2 = \pi (4 \times 10^{-5})^2 = 16\pi \times 10^{-10}\ \text{m}^2$
Strain $= \frac{T}{AY} = \frac{80/3}{16\pi \times 10^{-10} \times 2 \times 10^{11}} = \frac{80/3}{320\pi} = \frac{1}{12\pi}$
$\alpha = 12$

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