Question

The viscous force 'F' acting on a small sphere of radius 'r' moving with velocity 'v' through a liquid is given by F = 6πɳrv. Calculate dimensions of ɳ.

Answer 1

Answer:

The dimension formula of η is $[ML^{-1}T^{-1}]$.

Explanation:

Given that,

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)

$\eta=\dfrac{F}{krv}$

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

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

Hence, The dimension formula of η is $[ML^{-1}T^{-1}]$.

Answer 2

Explanation:

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