Verified Solution • 2025–26 Session moderate

QJEEM18Ph8e90: JEE Main 2018 Physics - Class 11 JEEM18Ph8e90 Solution: A disc rotates about its axis of symmetry in a horizontal plane …

A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of 3.53.5 revolutions per second. A coin placed at a distnce of 1.25 cm from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is : (g = 10 m/s2)

Multiple Choice Options

  • A 0.5
  • B 0.3
  • C 0.7
  • D 0.6

Step-by-Step Verified Solution

We have

mω2r=μmgm{\omega ^2}r = \mu mg ...... (1)

Given : rate of rotation = 3.5 rev/s

⇒\Rightarrow 1 revolution = 2 π\pi rad

JEE Main 2018 (Online) 15th April Evening Slot Physics - Circular Motion Question 61 English Explanation

That is, 3.5 revolutions = 3.5 ×\times 2 π\pi rad

Therefore, ω\omega = 3.5 ×\times 2 π\pi rad/s

r = 1.25 cm = 1.25 ×\times 10 −- 2 m

Thus, from Eq. (1), we have

mω2r=μmg⇒ω2r=μgm{\omega ^2}r = \mu mg \Rightarrow {\omega ^2}r = \mu g

⇒μ=ω2rg=(3.5×2π)2(1.25×10−2)10\Rightarrow \mu = {{{\omega ^2}r} \over g} = {{{{(3.5 \times 2\pi )}^2}(1.25 \times {{10}^{ - 2}})} \over {10}}

⇒μ=0.60\Rightarrow \mu = 0.60

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