Q

# the fundamental unit which has the same power in the dimensional formula of surface tension and viscosity? explain

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it is known that,
the formula of Tension = force/Area
So, the dimensional formula = $\frac{[MLT{^-2}]}{[L^2]} = ML^{-1}T^{-2}$

the formula of viscosity = $\nu = \frac{F}{A.(dv/dx)}$
Therefore, the dimensional formula
$\frac{[MLT^{-2}]}{[L^2][LT^{-2}]}=ML^{-2}T^0$

Hence Mass (M) has the same power.

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