Question

The coefficient of viscosity ‘η ‘ is defined according, f = 6 π η r v , where f is the friction force, r is a radius of the body and v is its velocity, then the dimensions of η is

A) \([M][L]^{-1}[T]^{-1}\)
B) \([M][L][T]^{-1}\)
C) \([M][L]^{-1}[T]\)
D) \([M][L]^2[T]^{-1}\)

Answer 1

The dimensions of the coefficient of viscosity 'η' derived from Stoke's law are [M][L]-1[T]-1, corresponding to Option A.

Step-by-step explanation:

The coefficient of viscosity 'η' has dimensions which can be derived from Stoke's law, Fs = 6πηrv, where Fs is the friction force, r is the radius of the body, and v is its velocity. Using dimensions, force (F) has the dimension [MLT-2], radius (r) has the dimension [L], and velocity (v) has the dimension [LT-1]. Solving for the dimensions of viscosity (η), we get:

η = F / (6πrv)

Therefore, the dimensions of η are:

[MLT-2] / ([L][LT-1]) = [ML-1T-1]

Hence, the correct answer is Option A) [M][L]-1[T]-1.