Coefficient of friction, μ =0.15
Inthe given situation, the coin having a force of friction greater than or equalto the centripetal force provided by the rotation of the disc will revolve withthe disc. If this is not the case, then the coin will slip from the disc.
Coin placed at 4 cm:
Radius of revolution, r‘ = 4 cm= 0.04 m
Angular frequency, ω = 2πν
Frictional force, f = μmg= 0.15 × m × 10 = 1.5m N
Centripetalforce on the coin:
Fcent.
= 0.49m N
Since f > Fcent,the coin will revolve along with the record.
Coin placed at 14 cm:
Radius, 
= 14 cm = 0.14 m
Angular frequency, ω = 3.49s–1
Frictional force, f‘ = 1.5m N
Centripetalforce is given as:
Fcent.
= m × 0.14 × (3.49)2
= 1.7m N
Since f < Fcent.,the coin will slip from the surface of the record.