Answer:
\[[F]=[ML{{T}^{-2}}]\] \[[6\pi \eta r\upsilon ]=[M{{L}^{-1}}{{T}^{-1}}][L][{{T}^{-1}}]=[ML{{T}^{-2}}]\] \[\therefore \] Dimensions of LHS = Dimensions of RHS. Hence the given formula for viscous force F is dimensionally correct.
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