JEE Main 2022 · 28 July · evening · Single correct · +4 / −1
A uniform metal chain of mass m and length 'L' passes over a massless and frictionless pulley. It is released from rest with a part of its length 'l' is hanging on one side and rest of its length 'L-l' is hanging on the other side of the pully. At a certain point of time, when l=L/x, the acceleration of the chain is g/2. The value of x is __________.
- (a) 6
- (b) 2
- (c) 1.5
- (d) 4
Answer and solution
Answer (d) 4
Solution
Physics sir · 4 steps
Only the extra hanging length pulls; the whole chain's mass has to be accelerated by it.
- Find the driving force. The chain has mass per unit length λ = m/L. The weights of equal lengths on the two sides balance. Only the extra length (L-l) - l = L - 2l on the longer side is left unbalanced, and its weight is m/L(L-2l)g.
- Apply Newton's law. This net force moves the entire chain, of mass m, with one common acceleration a. So m a = m/L(L-2l)g.
a = (L-2l)/L g
- Use the given acceleration. Put a = g/2: (L-2l)/L = 1/2 gives L - 2l = L/2, so l = L/4.
- So. Comparing l = L/4 with l = L/x gives x = 4, which is option (d).
The idea to keep. For a chain over a pulley, the net force is the weight of the extra length on the longer side. Divide it by the total mass of the chain, not by the mass of one side.
Answer from the NTA final answer key. Solution by PrepWiser.