Atwood Machine And Pulleys

Per NCERT Class 11, Chapter 5 (Laws of Motion), for an ideal Atwood machine with a massless string over a frictionless pulley, both masses share the same magnitude of acceleration because the string is inextensible. Writing Newton's Second Law for each mass and adding the equations eliminates the tension, giving the standard result a = (m₁ − m₂) g / (m₁ + m₂). Substituting, a = (6 − 4) × 10 / (6 + 4) = 20 / 10 = 2 m/s². The heavier 6 kg mass descends while the 4 kg mass rises with this acceleration. Option 4 m/s² uses only the mass difference times g divided by a single mass, mishandling the denominator. Option 1 m/s² halves the correct value. Option 10 m/s² wrongly assumes free fall, ignoring that the lighter mass holds back the system. It is instructive to see that if the two masses were equal, the numerator would vanish and the acceleration would be zero, leaving the system in balance regardless of how heavy the masses are. The pulley simply redirects the string tension and, being ideal, neither adds nor removes energy from the system. Plausibility check: the acceleration must be much smaller than g because the masses nearly balance, and 2 m/s² is a small fraction of 10 m/s², which is physically sensible for a near-balanced pulley.