Question

Test if the following equations are dimensionally correct:
(a) $h=\frac{2S\mathrm{cos\theta }}{\rho rg},$

(b) $v=\sqrt{\frac{P}{\rho }},$

(c) $V=\frac{\pi P{r}^{4}t}{8\eta l}$

(d) $v=\frac{1}{2\pi }\sqrt{\frac{mgl}{I}};$
where h = height, S = surface tension, $\rho$ = density, P = pressure, V = volume, $\eta =$ coefficient of viscosity, v = frequency and I = moment of interia.

Answer 1

(a) $h=\frac{2\mathrm{S}\cos \theta }{\mathrm{\rho }rg}$
Height, [h] = [L]
Surface Tension, $\left[\mathit{S}\right]=\frac{\left[F\right]}{\left[L\right]}=\frac{\left[{\mathrm{MLT}}^{-2}\right]}{\left[\mathrm{L}\right]}=\left[{\mathrm{MT}}^{-2}\right]$
Density, $\left[\rho \right]=\frac{\left[M\right]}{\left[\mathrm{I}\right]}=\left[{\mathrm{ML}}^{-3}{\mathrm{T}}^0\right]$
Radius, [r] = [L], [g]= [LT⁻²]
Now, $\frac{2\left[S\right]\cos \theta }{\left[\rho \right]\left[r\right]\left[g\right]}=\frac{\left[{\mathrm{MT}}^{-2}\right]}{\left[{\mathrm{ML}}^{-3}{\mathrm{T}}^0\right]\left[\mathrm{L}\right]\left[{\mathrm{LT}}^{-2}\right]}=\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right]=\left[\mathrm{L}\right]$

Since the dimensions of both sides are the same, the equation is dimensionally correct.

(b) $\nu =\sqrt{\left(\frac{\mathit{P}}{\mathit{\rho }}\right)}$
Velcocity, [ν] = [LT⁻¹]
Pressure, $P=\frac{\left[F\right]}{\left[A\right]}=\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right]$
Density, $\left[\rho \right]=\frac{\left[M\right]}{\left[V\right]}=\left[{\mathrm{ML}}^{-3}{\mathrm{T}}^0\right]$
Now, $\sqrt{\frac{\mathit{P}}{\mathit{\rho }}}={\left[\frac{\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right]}{\left[{\mathrm{ML}}^{-3}\right]}\right]}^{\frac{1}{2}}={\left[{\mathrm{L}}^2{\mathrm{T}}^{-2}\right]}^{1/2}=\left[{\mathrm{LT}}^{-1}\right]$
Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.

(c) $V=\frac{\left(\pi \mathit{}P\mathit{}{r}^{4}t\right)}{\left(8\eta l\right)}$
Volume, [V] = [L³]
Pressure, $P=\frac{\left[F\right]}{\left[A\right]}=\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right]$
[r]= [L] and [t] = [T]
Coefficient of viscosity,
$\left[\eta \right]=\frac{\left[F\right]}{6\mathrm{\pi }\left[\mathrm{r}\right]\left[\mathrm{v}\right]}=\frac{\left[{\mathrm{MLT}}^{-2}\right]}{\left[\mathrm{L}\right]\left[{\mathrm{LT}}^{-1}\right]}=\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-1}\right]$

Now, $\frac{\pi \left[P\right]{\left[r\right]}^4\left[t\right]}{8\left[\mathrm{\eta }\right]\left[l\right]}=\frac{\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right]\left[{\mathrm{L}}^4\right]\left[\mathrm{T}\right]}{\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-1}\right]\left[\mathrm{L}\right]}=\left[{\mathrm{L}}^3\right]$
Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.

(d) $\mathrm{\nu }=\frac{1}{2\pi }\sqrt{\left(\frac{mgl}{I}\right)}$
Frequency, ν = [T⁻¹]

$$\begin{gathered} \sqrt{\left(\frac{mgl}{I}\right)}=\sqrt{\frac{\left[\mathrm{M}\right]\left[{\mathrm{LT}}^{-2}\right]\left[\mathrm{L}\right]}{\left[{\mathrm{ML}}^2\right]}} \\ \Rightarrow {\left[\frac{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]}{\left[{\mathrm{ML}}^2\right]}\right]}^{\frac{1}{2}}=\left[{\mathrm{T}}^{-1}\right] \end{gathered}$$

Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.