Which of the following has the dimension of $M L^{0} T^{- 2}$ ?

Coefficient of viscosity

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Surface tension

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Vapour pressure

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Kinetic energy

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Solution

The correct option is A Surface tension
Surface tension, $γ = \frac{δ w}{d A} = \frac{F . d x}{d A}$

$= F L^{- 1}$

$= [M L T^{- 2}] [L^{- 1}]$

$= M L^{0} T^{- 2}$

Hence, option B is correct.

Suggest Corrections

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[M^{1} L^{2} T^{- 2}]

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P = \frac{a - t^{2}}{b x}

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