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NEET PHYSICSLAWS OF MOTIONMedium

Question

A uniform rope of length ll lies on a table. If the coefficient of friction is μ\mu, then the maximum length l1l_1 of the part of this rope which can overhang from the edge of the table without sliding down is:

A

μl\mu l

B

lμ+1\frac{l}{\mu + 1}

C

μl1+μ\frac{\mu l}{1 + \mu}

D

μlμ1\frac{\mu l}{\mu - 1}

Step-by-Step Solution

  1. System Definition: Let the total length of the rope be ll and its linear mass density (mass per unit length) be λ\lambda. Let l1l_1 be the length hanging over the edge and l2=ll1l_2 = l - l_1 be the length lying on the table.
  2. Forces:
  • Driving Force: The weight of the hanging portion tries to pull the rope down. Fdriving=mhangingg=(λl1)gF_{driving} = m_{hanging}g = (\lambda l_1)g.
  • Resisting Force: The static friction acting on the portion of the rope on the table opposes the motion. The normal force NN is equal to the weight of the portion on the table: N=mtableg=λ(ll1)gN = m_{table}g = \lambda (l - l_1)g. The maximum static friction is fmax=μN=μλ(ll1)gf_{max} = \mu N = \mu \lambda (l - l_1)g [Source 64].
  1. Equilibrium Condition: For the rope to be on the verge of sliding (maximum overhang), the driving force must equal the maximum static friction: (λl1)g=μλ(ll1)g(\lambda l_1)g = \mu \lambda (l - l_1)g
  2. Solving for l1l_1: Cancel λg\lambda g from both sides: l1=μ(ll1)l_1 = \mu (l - l_1) l1=μlμl1l_1 = \mu l - \mu l_1 l1(1+μ)=μll_1(1 + \mu ) = \mu l l1=μl1+μl_1 = \frac{\mu l}{1 + \mu }

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 MOTIONuniformlengthcoefficientfrictionmaximum

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