Question

F=6πηv^2r dimensionally correct or not?

I solved it already and I think it is dimensionally incorrect.
MLT^2≠ML^2T^-3
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Answer 1

Answer:

yes it's dimentionally incorrect. And you are right✔. F=MLT^-2 and it's dimension is ML^2T^-3

Explanation:

n=ML^-1T^-1

v^2= L^2T^-2

R=L

AND 6 and π are dimension less