KVPY SX SB Exam 19-Nov-2017 Question Paper

Question : 27 of 120

Marks: +1, -0

A uniform rope of total length l is at rest on a table with fraction f of its length hanging (see figure). If the coefficient of friction between the table and the chain is µ then

$f = µ$
$f = 1/ (1 + µ)$
$f = 1/ (1 + 1/µ)$
$f = 1/ (µ + 1/µ)$

Validate

Solution:

\f_r = µN = µM"g = µ (M -M_\f)g

at equilibrium.

\f_r = M'g

µM( 1 -\f)g = M_Fg

µ(1-\f) = \f

µ = (µ + 1) \f

\f=1/(1+1/µ)

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