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/(α π). The value of α is _________. [Use g=10 m / s²]

Solution

Physics sir · 4 steps

The blocks accelerate, so the tension is not either weight: find T first, then strain is stress over Y.

1. Tension in the wire. The 4 kg block moves down and the 2 kg block moves up with the same acceleration. Solving Newton's second law for both blocks gives the standard Atwood result for tension.

   T = (2m₁ m₂ g)/(m₁ + m₂) = (2 × 2 × 4 × 10)/6 = 80/3 N

2. Stress. The cross-section area is A = π r² = π × (4.0 × 10⁻⁵)² = 1.6π × 10⁻⁹ m². So the stress is T/A = (80/3)/(1.6π × 10⁻⁹) = (5 × 10¹⁰)/3π N m⁻².
3. Strain. Strain is stress divided by Young's modulus: (5 × 10¹⁰)/(3π × 2.0 × 10¹¹) = 1/12π.
4. So. Comparing with 1/απ gives α = 12.

The common slip: 8. Taking the tension as the weight of the 4 kg block, 40 N, ignores the acceleration and gives a strain of 1/8π, so 8 instead of 12.

The idea to keep. Whenever a wire holds accelerating masses, get the tension from Newton's laws first. Only then use stress = T/A and strain = stress/Y.

Answer from the NTA final answer key. Solution by PrepWiser.