check if it is dimensionally correct f=6πηrv
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Answer:
f = force [MLT^-2]
η = coefficient of viscosity [ML^-1T^-1]
r= length [L]
v = velocity [LT^-2]
f = 6πηrv
RHS :
6πηrv
=[MLT^-2] [L] [LT^-1] {some constant
like π and integers
are dimensionaless
=[MLT^-2]
LHS :
f
=[MLT^-2]
hence LHS = RHS.
so, the equation is dimensionaly correct.
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