Turpentine oil is flowing through a capillary 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- -pr2-x24vl- Here v is velocity of oil at a distance x from the axis of the tube- From this relation- the dimensional formula of - is

The correct option is D [ML^ 1T^ 1] [η ]= [p] [ r ^ 2 x ^ 2 ] [v] ^ 1 [l] ^ 1 #160; #160; #160; = M L ^ 1 T ^ 2 L ^ 2 L ^ ...

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Standard XII

Physics

Dimensional Analysis

Turpentine oi...

Question

Turpentine oil is flowing through a capillary 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: $η = \frac{p (r^{2} - x^{2})}{4 v l}$ . Here v is velocity of oil at a distance $x$ from the axis of the tube. From this relation, the dimensional formula of $η$ is

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

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[M L T^{- 1}]

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

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

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Solution

The correct option is D $[M L^{- 1} T^{- 1}]$
$[η] = [p] [r^{2} - x^{2}] {[v]}^{- 1} {[l]}^{- 1}$
$= M L^{- 1} T^{- 2} L^{2} L^{- 1} T L^{- 1}$
$= M L^{- 1} T^{- 1}$

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The correct option is D [ML−1T−1][η]=[p][r2−x2][v]−1[l]−1 =ML−1T−2L2L−1TL−1 =ML−1T−1

The correct option is D $[M L^{- 1} T^{- 1}]$
$[η] = [p] [r^{2} - x^{2}] {[v]}^{- 1} {[l]}^{- 1}$
$= M L^{- 1} T^{- 2} L^{2} L^{- 1} T L^{- 1}$
$= M L^{- 1} T^{- 1}$