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


ANSWER

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


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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