Q.

A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of 3.5 revolutions per second. A coin placed at a distance of 1.25cm from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is (g=10m/s2)

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a

0.5

b

0.7

c

0.3

d

0.6

answer is D.

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Detailed Solution

Using, μmg=mv2r=mrω2  ω=2πn=2π×3.5=7π rad/secRadius, r=1.25 cm=1.25×10−2mCoefficient of friction, μ=? μmg=m(rω)2r (∵ν=rω)μmg=mrω2⇒μ=rω2g=1.25×10−2×(7×227)210μ=1.25×10−2×22210=0.6
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