Determine if the following equation is dimensionally correct:
$h = \dfrac{2 S \cos\theta}{\rho r g}$
where $h =$ height, $S =$ surface tension, $\rho =$ density, $P =$ pressure, $V =$ volume, $\eta =$ coefficient of viscosity, $\nu =$ frequency and $I =$ moment of inertia.

Question

Determine if the following equation is dimensionally correct:

h = \dfrac{2 S \cos\theta}{\rho r g}

where

h =

height,

S =

surface tension,

\rho =

density,

P =

pressure,

V =

volume,

\eta =

coefficient of viscosity,

\nu =

frequency and

I =

moment of inertia.

Quantrex Academy · Accepted Answer

Correct answer: D. The equation is dimensionally correct.

Dimensions of the left-hand side: [h] = [L] Dimensions of the quantities on the right-hand side: [S] = [M T^ -2 ] [ θ] = [M^0 L^0 T^0] [ ] = [M L^ -3 ] [r] = [L] [g] = [L T^ -2 ] Substituting these dimensions into the right-hand side: RHS = [M T^ -2 ] [M L^ -3 ] [L] [L T^ -2 ] RHS = [M T^ -2 ] [M L^ -1 T^ -2 ] RHS = [L] Since the dimensions of the left-hand side equal the dimensions of the right-hand side, the equation is dimensionally correct.

Determine if the following equation is dimensionally correct: h =...

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