Determine if the following equation is dimensionally correct:
$\nu = \dfrac{1}{2 \pi} \sqrt{\dfrac{mgl}{I}}$
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:

\nu = \dfrac{1}{2 \pi} \sqrt{\dfrac{mgl}{I}}

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: B. The equation is dimensionally correct.

Dimensions of the left hand side (LHS):

[\nu] = [T^{-1}]

Dimensions of the terms on the right hand side (RHS):

[m] = [M]

[g] = [L T^{-2}]

[l] = [L]

[I] = [M L^2]

Substituting these into the expression under the square root:
\left[\dfrac{mgl}{I}\right] = \dfrac{[M][L T^{-2}][L]}{[M L^2]} = [T^{-2}]

<br><br>Taking the square root for the RHS:<br>\left[\sqrt{\dfrac{mgl}{I}}\right] = \sqrt{[T^{-2}]} = [T^{-1}]

Since the dimensions of the LHS and RHS are equal, the equation is dimensionally correct.

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