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 $(g = 10 m / s^{2})$

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

0.5

0.7

0.6

0.3

answer is D.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Using, $μ m g = \frac{m v^{2}}{r} = m r ω^{2}$

$ω = 2 π n = 2 π \times 3.5 = 7 π r a d / \sec$

Radius, $r = 1.25 c m = 1.25 \times 10^{- 2} m$

Coefficient of friction, $μ = ?$

$μ m g = \frac{m {(r ω)}^{2}}{r} (∵ ν = r ω)$

$μ m g = m r ω^{2}$

$\Rightarrow μ = \frac{r ω^{2}}{g} = \frac{1.25 \times 10^{- 2} \times {(7 \times \frac{22}{7})}^{2}}{10}$

$μ = \frac{1.25 \times 10^{- 2} \times 22^{2}}{10} = 0.6$

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test