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Q.

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 the oil is given by $η = \frac{P (r^{2} - x^{2})}{4 v l}$ where $v$ is the velocity of oil at a distance x from the axis of the tube. From this relation, the dimensions of viscosity $η$ are

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a

$[M^{0} L^{0} T^{0}]$

b

$[M L^{2} T^{- 2}]$

c

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

d

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

answer is D.

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Detailed Solution

detailed_solution_thumbnail

$η= \frac{p. (r^{2} {-x}^{2})}{4VL}$

For the dimension of viscosity $η= \frac{[pressure] [L^{2}]}{[{LT}^{-1}] [L]}$

We have to know the dimensions of pressure; which in turn is force per unit area.

$[pressure] = [{ML}^{-1} T^{-2}]$

$η= \frac{[{ML}^{-1} T^{-2}] [L^{2}]}{[{LT}^{-1}] [L]} = [{ML}^{-1} T^{-1}]$

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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 the oil is given by η=P(r2−x2)4vl where v is the velocity of oil at a distance x from the axis of the tube. From this relation, the dimensions of viscosity η are