2.A skater weighing 1000N skates at a speed of 20m/s on ice maintained...
Solution:
Given data:
Weight of the skater, W = 1000 N
Speed of the skater, v = 20 m/s
Area of contact, A = 0.001 m²
Coefficient of friction, μ = 0.02
Viscosity of water, η = 0.01 Ns/m²
We need to find the average thickness of a film of water existing at the interface between the skater and ice.
Formula used:
The force of friction between the skater and ice is given by:
f = μW
The force of viscosity acting on the water film is given by:
F = ηAv/d
where d is the thickness of the water film.
At equilibrium, these two forces are balanced, i.e.,
μW = ηAv/d
d = ηAv/μW
Calculation:
Substituting the given values in the above expression, we get:
d = (0.01 Ns/m²) × (0.001 m²) × (20 m/s) / (0.02 × 1000 N)
d = 10^-6 m
Therefore, the average thickness of the water film existing at the interface between the skater and ice is 10^-6 m.
Answer: Option (b) 10^-6m