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NEET PHYSICSLAWS OF MOTIONMedium

Question

A heavy uniform chain lies on a horizontal table-top. If the coefficient of friction between the chain and 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:

A

20%

B

25%

C

35%

D

15%

Step-by-Step Solution

  1. System Analysis: Let the total length of the chain be LL and its total mass be MM. Let the length of the chain hanging over the edge be ll. The length of the chain remaining on the table is L−lL - l.
  2. Forces Involved:
  • Driving Force: The weight of the hanging portion pulls the chain down. Mass of hanging part = (l/L)M(l/L)M. Force Fg=(l/L)MgF_g = (l/L)Mg.
  • Resisting Force: The limiting static friction acting on the portion on the table balances the pulling force. Mass on table = ((L−l)/L)M((L-l)/L)M. Normal reaction N=((L−l)/L)MgN = ((L-l)/L)Mg. Friction f=μN=μ((L−l)/L)Mgf = \mu N = \mu ((L-l)/L)Mg [Source 74].
  1. Equilibrium Condition: For the maximum length to hang without sliding, the driving force must equal the limiting friction: lLMg=μ(L−lL)Mg\frac{l}{L}Mg = \mu \left(\frac{L-l}{L}\right)Mg Canceling common terms (Mg/LMg/L): l=μ(L−l)l = \mu(L - l) l=μL−μll = \mu L - \mu l l(1+μ)=μLl(1 + \mu) = \mu L lL=μ1+μ\frac{l}{L} = \frac{\mu}{1 + \mu}
  2. Calculation: Given μ=0.25\mu = 0.25: Fraction=0.251+0.25=0.251.25=15=0.2\text{Fraction} = \frac{0.25}{1 + 0.25} = \frac{0.25}{1.25} = \frac{1}{5} = 0.2 Percentage=0.2×100%=20%\text{Percentage} = 0.2 \times 100\% = 20\%

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from LAWS OF MOTION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSLAWS OF MOTIONuniformhorizontaltabletopcoefficientfriction

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