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Viscosity

The dimension...

Question

The dimensions of velocity gradient are:

[

T^{- 1}

]

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[

T

]

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[

T^{2}

]

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[

T^{\circ}

]

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Solution

The correct option is A [ $T^{- 1}$ ]

Velocity gradient $= \frac{d v}{d x}$
so dimensions of velocity gradient is $\frac{[L T^{- 1}]}{[L]} = [T^{- 1}]$

Suggest Corrections

Similar questions

t^{2} + t + 1 = 0,

(t + \frac{1}{t}) + (t^{2} + \frac{1}{t^{2}}) + \dots + (t^{27} + \frac{1}{t^{^{27}}})

\frac{α}{β}

F = \frac{α - t^{2}}{β v^{2}}

F

v

t

x = t^{2} + t + 1

y = t^{2} - t + 1

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