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 -pr2-x24vl- where v is the velocity of oil at a distance x from the axis of the tube- From this relation-the dimensions of viscosity - 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 ] [ ...

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Byju's Answer

Standard XII

Physics

Dimensional Analysis

Turpentine oi...

Question

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 $η$ is:

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

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[M L T^{- 1}]

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

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

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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−2][L2][L−1T][L−1] =ML−1T−1

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