Question

find an expression for viscous force f acting on a tiny steel ball of radius r moving in a viscous liquid of viscosity n with a constant speed v by the method of dimensional analysis​

Answer 1

Answer:

The viscous force F = 6\pi \eta rv

Let, F = k\eta rv

\eta=\dfrac{F}{krv}....(I)

Here, k=6\pi= dimensionless constant

We know that,

F= [MLT^{-2}]

r=[L]

v=[LT^{-1}]

Put the dimension of all element in equation (I)

\eta=\dfrac{F}{krv}

\eta=\dfrac{ [MLT^{-2}]}{[L][LT^{-1}]}

\eta=[ML^{-1}T^{-1}]

Answer 2

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