Laws of Motion

JEE Main level, single correct, medium

22 of 210

Question 22 of 210

A uniform metal chain of mass mm and length LL passes over a massless and frictionless pulley. It is released from rest with a part of its length ll is hanging on one side and rest of its length L−lL-l is hanging on the other side of the pulley. At a certain point of time, when l=Lxl=\dfrac{L}{x}, the acceleration of the chain is g2\dfrac{g}{2}. The value of xx is _____.

Show answer and solution

Answer: (D) 44


According to the information given in the question, the part of the chain that envelopes the pulley is very small. Therefore, the acceleration of the chain can be written as,

a=(m2−m1)(m2+m1)ga=\dfrac{\left({m}_{2}-{m}_{1}\right)}{\left({m}_{2}+{m}_{1}\right)}g
Since
a=g2a=\dfrac{g}{2}
g2=(m2−m1)(m2+m1)g⇒g2=(L−l)−lLg⇒l=L4=Lx\dfrac{g}{2}=\dfrac{\left({m}_{2}-{m}_{1}\right)}{\left({m}_{2}+{m}_{1}\right)}g \Rightarrow \dfrac{g}{2}=\dfrac{\left(L-l\right)-l}{L}g \Rightarrow l=\dfrac{L}{4}=\dfrac{L}{x}
⇒x=4\Rightarrow x=4

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Laws of motion, friction and pulleys simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Laws of Motion (Newton's laws of motion), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.