If velocity ' $V$ ', kinetic energy ' $k$ ' and time ' $T$ ' are taken as fundamental quantities, then the dimensional formula of surface tension is:

{K V}^{2} T^{2}

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{K V}^{- 2} T^{- 2}

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{K^{2} V}^{2} T^{- 2}

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{K^{2} V}^{- 2} T^{- 2}

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Solution

The correct option is B ${K V}^{- 2} T^{- 2}$
$S u r f a c e T e n s i o n = \frac{- f}{l} = \frac{M^{1} L^{1} T^{- 2}}{L^{1}} = {M T}^{- 2}$
${M T}^{- 2} = K^{α} V^{β} T^{γ}$
${M T}^{- 2} = {({M L}^{2} T^{- 2})}^{α} {({L T}^{- 1})}^{β} {(T)}^{γ}$
$M^{1} T^{- 2} = M^{α} L^{2 α + β} T^{- 2 α - β + γ}$
$α = 1, 2 α + β = 0 \Rightarrow β = - 2$
$- 2 α - β + γ = - 2$
$γ = - 2$
$\Rightarrow {M T}^{- 2} = S u r f a c e T e n s i o n = K^{1} V^{- 2} T^{- 2}$

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If velocity 'V', kinetic energy 'k' and time 'T' are taken as fundamental quantities, then the dimensional formula of surface tension is:

The correct option is B KV−2T−2Surface Tension =−fl=M1L1T−2L1=MT−2MT−2=KαVβTγMT−2=(ML2T−2)α(LT−1)β(T)γM1T−2=MαL2α+βT−2α−β+γα=1,2α+β=0⇒β=−2−2α−β+γ=−2γ=−2 ⇒MT−2=Surface Tension=K1V−2T−2

If velocity ' $V$ ', kinetic energy ' $k$ ' and time ' $T$ ' are taken as fundamental quantities, then the dimensional formula of surface tension is: