A disc rotates about its aixs of symmetry in a horizontal plane at a steady rate of 3.5 revolutions per second A coin placed at a distance fo 1.25 cm form the axis of ratation remains at rest on the disc The coefficient of friction between the coin and the disc is : `(g=10//s^(2))`

Options

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

Answer

D

Explanation

Using `mu mg =(mv^(2))/(r)= mr omega^(2)`
`omega= 2pi n= 2pi xx3.5= 7 pi"rad"//sec`
Radius `r=1.25 cm =1.25 xx10^(-2)m`
coefficient of friction `mu=?`
`rArr =(r omega^(2))/(g(1.25xx10^(-2)xx(7xx(22)/(7))^(2))/(10)`
`=(1.25xx10^(2)xx22^(2))/(10)=0.6`