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 by η = (P(r² - x²))/(4vl) where v is the velocity of oil at a distance x from the axis of the tube. The dimensions of η are (a) [M⁰L⁰T⁰] (b) [MLT⁻¹] (c) [ML²T⁻²] (d) [ML⁻¹T⁻¹]

Answer: (d). [r² - x²] = L², [P] = ML⁻¹T⁻², [v] = LT⁻¹, [l] = L. [η] = ML⁻¹T⁻²· L²LT⁻¹· L. [η] = ML⁻¹T⁻¹, so the answer is (d).

Turpentine oil is flowing through a tube of length l and radius r… | NEET 1993 (AIPMT)

Q82NEET 1993 · AIPMTMedium

Applications of Dimensional Analysis

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 by $\eta = \frac{P(r^2 - x^2)}{4vl}$ where $v$ is the velocity of oil at a distance $x$ from the axis of the tube. The dimensions of $\eta$ are

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Answer: (d)

$[r^2 - x^2] = L^2$ , $[P] = ML^{-1}T^{-2}$ , $[v] = LT^{-1}$ , $[l] = L$ .
$[\eta] = \dfrac{ML^{-1}T^{-2}\cdot L^2}{LT^{-1}\cdot L}$ .
$[\eta] = ML^{-1}T^{-1}$ , so the answer is (d).

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