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 u is the velocity of oil at a distance x from the axis ofthe tube. From this relation, the dimensions of viscosity η are :

η=p(r2 - x2)4vl=pr2vl=prv (ML-1 T2 )LT-1=ML-1T-1

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 u is the velocity of oil at a distance x from the axis ofthe tube. From this relation, the dimensions of viscosity η are :

η=p(r2 - x2)4vl=pr2vl=prv (ML-1 T2 )LT-1=ML-1T-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 : $η = \frac{P (r^{2} - x^{2})}{4 vl}$ where u 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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[M⁰L⁰T⁰]

[MLT^-1]

[ML²T^-2]

[ML^-1T^-1]

answer is D.

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Detailed Solution

$η = [\frac{p (r^{2} - x^{2})}{4 vl}] = \frac{{pr}^{2}}{vl} = \frac{pr}{v} [\frac{({ML}^{- 1} T 2)}{{LT}^{- 1}}] = [{ML}^{- 1} T^{- 1}]$

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