A light string passing over a smooth light pulley connects two blocks of masses m₁ and m₂( where m₂>m₁). If the acceleration of the system is g/√2, then the ratio of the masses m₁/m₂ is:

Solution

Physics sir · 4 steps

Write the Atwood acceleration, set it equal to g/√2, and solve for m₁/m₂.

1. Atwood acceleration. The heavier block m₂ goes down and m₁ goes up. Writing Newton's second law for each block and adding the two equations removes the tension T.

   a = ((m₂ - m₁) g)/(m₁ + m₂) = g/√2

2. Use the ratio. Divide the top and bottom by m₂ and let r = m₁/m₂. This gives (1 - r)/(1 + r) = 1/√2, so √2(1 - r) = 1 + r.
3. Solve for r. Collect the r terms: √2 - 1 = r(√2 + 1), so r = (√2 - 1)/(√2 + 1) ≈ 0.17. Check: (1 - 0.17)/(1 + 0.17) ≈ 0.707 = 1/√2.
4. So. m₁/m₂ = (√2 - 1)/(√2 + 1), which is option (c). It is also the only option less than 1, as it must be when m₂ > m₁.

The idea to keep. For two blocks over a pulley, a = (m₂ - m₁)g/(m₁ + m₂). Divide through by one mass to turn it into an equation in the ratio, then check that the ratio is less than 1.

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