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Dimensional Analysis

If energy E...

Question

If energy $(E)$ , velocity $(v)$ and time $(T)$ are fundamental units. What will be the dimension of surface tension?

[S] = E v^{- 2} T^{- 2}

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[S] = E v^{- 1} T^{- 2}

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[S] = E v^{- 2} T^{- 1}

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None of these

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Solution

The correct option is C $[S] = E v^{- 2} T^{- 2}$
Surface tension $(S)$ $= \frac{w o r k d o n e}{A r e a} = \frac{E n e r g y}{A r e a} = [\begin{matrix} \frac{E}{A} \end{matrix}] = [\begin{matrix} \frac{E}{L^{2}} \end{matrix}]$
$∵ L = v T ∴ S = \frac{E}{(v T)^{2}} = E v^{- 2} T^{- 2}$

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Similar questions

E

V

T

If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, thendimensional formula of surface tension will be :

^{'} K^{'},

^{'} v^{'}

^{'} T^{'}

S =

V

k

T

E = a [1 + c o s (2 π f_{1} t)] c o s (2 π f_{2} t)

f_{1} = 1.2 \times 10^{15} H z

f_{2} = 3.6 \times 10^{15} H z

h = 6.6 \times 10^{- 34} J s

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