Physics

# Turpentine oil is flowing through a tube of length l and radius r. The pressure difference between the two ends of the tube is p. The viscosity of oil is given $\displaystyle \eta = \frac{p(r^2 - x^2)}{4vl}$, where v is velocity of oil at a distance x from the axis of the tube. The dimensional formula of viscosity is

$[M^1 L^{-1} T^{-1}]$

##### SOLUTION
Dimensions of $P=[ML^{-1}T^{-2}]$

Dimentions of $r=[L]$

Dimentions of $v=[LT^{-1}]$

Dimensions of $I=[L]$

$\displaystyle \eta = \frac{p(r^2 - x^2)}{4vl}$

$= \dfrac{[M L^{-1} T^{-2}] [L^2]}{[LT^{-1}] [L]}$

$= [M L^{-1} T^{-1}]$

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Single Correct Medium Published on 18th 08, 2020
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