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Zorluk: OrtaDimensions of Physical Quantities and Dimensional Analysis

The viscous force FF acting on a small sphere of radius rr moving with velocity vv through a liquid is given by Stokes' law, F=6πηrvF = 6\pi \eta r v, where η\eta is the coefficient of viscosity. Which of the following expressions represents the base dimensions of η\eta?

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

Cevap

ML−1T−1M L^{-1} T^{-1}
Rearranging Stokes' law gives η=F6πrv\eta = \frac{F}{6\pi r v}. Substituting base dimensions [F]=MLT−2[F] = M L T^{-2}, [r]=L[r] = L, and [v]=LT−1[v] = L T^{-1} yields [η]=MLT−2L2T−1=ML−1T−1[\eta] = \frac{M L T^{-2}}{L^2 T^{-1}} = M L^{-1} T^{-1}.

Adım Adım Çözüm

1
Express Stokes' law in terms of the coefficient of viscosity
η=F6πrv\eta = \frac{F}{6\pi r v}
Isolating η\eta allows substitution of fundamental dimensions.
2
Substitute fundamental dimensions for force, radius, and velocity
[η]=MLT−2L⋅LT−1[\eta] = \frac{M L T^{-2}}{L \cdot L T^{-1}}
The constant 6π6\pi is dimensionless, while [F]=MLT−2[F] = M L T^{-2}, [r]=L[r] = L, and [v]=LT−1[v] = L T^{-1}.
3
Simplify the powers of base quantities MM, LL, and TT
[η]=ML1−2T−2−(−1)=ML−1T−1[\eta] = M L^{1 - 2} T^{-2 - (-1)} = M L^{-1} T^{-1}
Applying algebraic rules of exponents simplifies the expression.

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Dimensions of Physical Quantities and Dimensional Analysis
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