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Q.Derive Stokes' law by dimensional analysis.

Assam AhsecAHSEC Higher Secondary (HS) 1st Year Examination 2025Subjective· 3mImportance★★★★★
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Dimensional analysis shows the viscous force on a sphere must be F = k·ηrv, and experiment/exact theory fixes k = 6π, giving Stokes' law F = 6πηrv.

Assume the viscous drag force FF on a small sphere moving slowly through a viscous fluid depends on: coefficient of viscosity η\eta, radius of the sphere rr, and speed vv:

F=k ηarbvcF = k\,\eta^{a}r^{b}v^{c} (k = dimensionless constant)

Dimensions: [F]=MLT−2[F]=MLT^{-2}, [η]=ML−1T−1[\eta]=ML^{-1}T^{-1}, [r]=L[r]=L, [v]=LT−1[v]=LT^{-1}.

MLT−2=(ML−1T−1)a(L)b(LT−1)c=Ma L(−a+b+c) T(−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:

M: 1=a⇒a=11 = a \Rightarrow a = 1

T: −2=−a−c⇒−2=−1−c⇒c=1-2 = -a - c \Rightarrow -2 = -1-c \Rightarrow c = 1

L: 1=−a+b+c⇒1=−1+b+1⇒b=11 = -a+b+c \Rightarrow 1 = -1+b+1 \Rightarrow b = 1

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