All surfaces shown in figure are assumed to be frictionless and the pulleys and the string are light. The acceleration of the block of mass 2 kg is :

Solution

Physics sir · 3 steps

The movable pulley halves the 4 kg block's motion, so the 2 kg block moves twice as fast.

1. String constraint. One string runs from the 2 kg block, over the top pulley, under the movable pulley and up to the ceiling. If the movable pulley drops by x, both strands around it lengthen by x, so the 2 kg block slides 2x up the incline. Take a as the 4 kg block's acceleration downward. The 2 kg block then has 2a up the incline.
2. Forces on each. On the 2 kg block, T pulls up the incline and 2g sin 30° = g pulls down it, so T - g = 2(2a). On the 4 kg block, two strands pull up with 2T and its weight 4g pulls down, so 4g - 2T = 4a, which gives T = 2g - 2a.

   (2g - 2a) - g = 4a ⇒ a = g/6

3. So. The 2 kg block's acceleration is 2a = 2 × g/6 = g/3, up the incline. The answer is (d).

The idea to keep. With a movable pulley, first write the string-length constraint to see how the accelerations are related. Then the block hanging from the pulley feels 2T while the free end feels T.

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