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Zorluk: OrtaPhysical Quantities, Units and Dimensions

The resistive force FF experienced by a small sphere of radius rr moving at velocity vv through a viscous fluid is given by Stokes' law: F=6πηrvF = 6\pi \eta r v, where η\eta is the coefficient of viscosity. What is the dimensional formula of η\eta?

  1. M L−1T−1\text{M L}^{-1}\text{T}^{-1}Cevap
  2. B
    M L T−1\text{M L T}^{-1}
  3. C
    M L−1T−2\text{M L}^{-1}\text{T}^{-2}
  4. D
    M L−2T−1\text{M L}^{-2}\text{T}^{-1}

Cevap

The dimensional formula of the coefficient of viscosity η\eta is M L−1T−1\text{M L}^{-1}\text{T}^{-1}.
Rearranging Stokes' law gives η=F6πrv\eta = \frac{F}{6\pi r v}. Since the constant 6π6\pi has no dimensions, substituting [F]=M L T−2[F] = \text{M L T}^{-2}, [r]=L[r] = \text{L}, and [v]=L T−1[v] = \text{L T}^{-1} yields [η]=M L T−2L2T−1=M L−1T−1[\eta] = \frac{\text{M L T}^{-2}}{\text{L}^2 \text{T}^{-1}} = \text{M L}^{-1}\text{T}^{-1}.

Adım Adım Çözüm

1
Isolate the coefficient of viscosity η\eta from Stokes' formula
η=F6πrv\eta = \frac{F}{6\pi r v}
Numerical constants like 6π6\pi are dimensionless, so [η]=[F][r][v][\eta] = \frac{[F]}{[r][v]}.
2
Substitute the fundamental dimensions for force, radius, and velocity
[F]=M L T−2[F] = \text{M L T}^{-2}, [r]=L[r] = \text{L}, and [v]=L T−1[v] = \text{L T}^{-1}
Force is mass times acceleration, radius is length, and velocity is displacement per unit time.
3
Perform exponent simplification for like base dimensions
[η]=M L T−2L⋅L T−1=M L T−2L2T−1=M L1−2T−2−(−1)=M L−1T−1[\eta] = \frac{\text{M L T}^{-2}}{\text{L} \cdot \text{L T}^{-1}} = \frac{\text{M L T}^{-2}}{\text{L}^2 \text{T}^{-1}} = \text{M L}^{1-2} \text{T}^{-2-(-1)} = \text{M L}^{-1}\text{T}^{-1}
Subtract indices of denominator dimensions from those in the numerator.

Anahtar Kavram

Dimensional Analysis of Viscosity

Alternatif Yöntem

Alternatively, unit derivation can be used: the SI unit of viscosity is N⋅s⋅m−2\text{N}\cdot\text{s}\cdot\text{m}^{-2}. Replacing Newtons with fundamental SI units (kg⋅m⋅s−2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) gives (kg⋅m⋅s−2)⋅s⋅m−2=kg⋅m−1⋅s−1(\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \cdot \text{s} \cdot \text{m}^{-2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-1}, which corresponds directly to [M L−1T−1][\text{M L}^{-1}\text{T}^{-1}].
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Physical Quantities, Units and Dimensions

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