Two blocks of masses m₁ and m₂ ( m₁ > m₂ ) are connected by a light inextensible string passing over a smooth fixed pulley. The system is released from rest. If the heavier mass m₁ falls a distance h in time t = 6h g , then the ratio of their masses m₁ m₂ is

Step-by-step solution

From the kinematic equation for uniformly accelerated motion starting from rest: h = 1 2 a t^2 Substitute the given time t = 6h g into the equation: h = 1 2 a ( 6h g ) Simplifying this gives: h = 3ah g a = g 3 For an Atwood machine, the acceleration of the system is given by: a = m₁ - m₂ m₁ + m₂ g Equating the two expressions for acceleration: m₁ - m₂ m₁ + m₂ g = g 3 m₁ - m₂ m₁ + m₂ = 1 3 Cross-multiplying yields: 3(m₁ - m₂) = m₁ + m₂ 3m₁ - 3m₂ = m₁ + m₂ 2m₁ = 4m₂ m₁ m₂ = 2 Answer: 2