Question 129: Units , Dimensions and Error Analysis

Question

If momentum [P], area $$[\mathrm{A}]$$ and time $$[\mathrm{T}]$$ are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is :

Accepted Answer

$$[\eta ] = [M{L^{ - 1}}{T^{ - 1}}]$$
Now if $$[\eta ] = {[P]^a}{[A]^b}{[T]^c}$$
$$ \Rightarrow [M{L^{ - 1}}{T^{ - 1}}] = {[M{L^1}{T^{ - 1}}]^a}{[{L^2}]^b}{[T]^c}$$
$$ \Rightarrow a = 1,\,a + 2b =  - 1,\, - a + c =  - 1$$
$$ \Rightarrow a = 1,\,b =  - 1,\,c = 0$$
$$ \Rightarrow [\eta ] = [P]{[A]^{ - 1}}{[T]^0}$$
$$ = [P{A^{ - 1}}{T^0}]$$

Answer

$$\left[\mathrm{P} \,\mathrm{A}^{-1} \mathrm{~T}^{0}\right]$$

Answer

$$\left[\mathrm{P} \,\mathrm{A}\mathrm{~T}^{-1}\right]$$

Answer

$$\left[\mathrm{P}\,\mathrm{A}^{-1} \mathrm{~T}\right]$$

Answer

$$\left[\mathrm{P} \,\mathrm{A}^{-1} \mathrm{~T}^{-1}\right]$$

Units , Dimensions and Error Analysis

If momentum [P], area $[\mathrm{A}] and time [\mathrm{T}]$ are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is :

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