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

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0.5

0.7

0.6

0.3

answer is D.

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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, $μ = ?$

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

$μ m g = m r ω^{2}$

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

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

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