StreakPeaked· Practice

ExamsJEE AdvancedPhysics

A block of mass 4M rests on a frictionless horizontal ice slab and is connected via a light inextensible string over a smooth fixed pulley to a hanging mass M. An external horizontal force of magnitude 2Mg is also applied to the 4M block in the direction that would accelerate it away from the pulley. Taking g as the acceleration due to gravity, the tension T in the connecting string is:

  1. T = 6Mg / 5
  2. T = 2Mg / 3
  3. T = 4Mg / 5
  4. T = Mg / 2

Correct answer: T = 6Mg / 5

Solution

The external force 2Mg accelerates the 4M block away from the pulley. By Newton's third law and the inextensible string, the hanging mass M rises with the same acceleration a. Setting up equations: (i) For M: T - Mg = Ma; (ii) For 4M: 2Mg - T = 4Ma. Adding gives Mg = 5Ma, so a = g/5. Substituting back: T = Mg + Ma = Mg + Mg/5 = 6Mg/5.

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →