Physics Circle home Student Login

Home›Chapters›Laws of Motion PYQs›Two-Body Friction›Question 13

NEET 1991 (AIPMT) · Two-Body Friction

Two-Body Friction question from NEET 1991 (AIPMT)

Q13NEET 1991 · AIPMTEasy

Two-Body Friction

A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of the table is

Show answer and solution

Answer: (a)

  1. Let a fraction ff hang. Weight of hanging part =f Mg= f\,Mg; maximum friction on the table part =μ(1−f)Mg= \mu(1-f)Mg.
  2. Limiting case: f=μ(1−f)⇒f=μ1+μf = \mu(1-f) \Rightarrow f = \frac{\mu}{1+\mu}
  3. f=0.251.25=0.2f = \frac{0.25}{1.25} = 0.2
  4. So the answer is (a) 20%.
Report a mistake

More questions on Two-Body Friction