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Q.Derive Stokes' law by dimensional analysis.

Assam AhsecAHSEC Higher Secondary (HS) 1st Year Examination 2018Subjective· 3mImportance★★★★★
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Assuming F=k ηarbvcF=k\,\eta^a r^b v^c and matching M, L, T powers on both sides fixes a=b=c=1a=b=c=1, giving Stokes' law F=6πηrvF=6\pi\eta rv (with k=6πk=6\pi from experiment/full theory).

When a small sphere of radius rr moves with speed vv through a viscous fluid of coefficient of viscosity η\eta, it experiences a retarding viscous force FF. Assume this force depends on η\eta, rr, and vv as a power-law relation:

F=k ηa rb vcF = k\,\eta^a\,r^b\,v^c

where kk is a dimensionless constant.

Dimensions of each quantity:

  • Force: [F]=[MLT−2][F] = [MLT^{-2}]
  • Coefficient of viscosity: [η]=[ML−1T−1][\eta] = [ML^{-1}T^{-1}]
  • Radius: [r]=[L][r] = [L]
  • Velocity: [v]=[LT−1][v] = [LT^{-1}]

Substituting into the assumed relation:

[MLT−2]=[ML−1T−1]a[L]b[LT−1]c=[MaL−a+b+cT−a−c][MLT^{-2}] = [ML^{-1}T^{-1}]^a[L]^b[LT^{-1}]^c = [M^a L^{-a+b+c} T^{-a-c}]

Equating powers on both sides:

  • Mass: a=1a = 1
  • Time: −a−c=−2⇒c=2−a=1-a-c = -2 \Rightarrow c = 2-a = 1 …

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