Test if the following equation is dimensionally correct:

\[h = \frac{2S cos\theta}{\text{ prg }},\]

where h = height, S = surface tension, ρ = density, I = moment of interia.

#### Solution

\[h = \frac{2S \cos \theta}{\text{ prg }}\]

Height, [h] = [L]

Surface Tension,

\[\left[ S \right] = \frac{\left[ F \right]}{\left[ L \right]} = \frac{\left[ {MLT}^{- 2} \right]}{\left[ L \right]} = \left[ {MT}^{- 2} \right]\]

Density,

\[\left[ \rho \right] = \frac{\left[ M \right]}{\left[ I \right]} = \left[ {ML}^{- 3} T^0 \right]\]

Radius, [r] = [L], [g]= [LT^{−2}]

Now,

\[\frac{2\left[ S \right]\cos \theta}{\left[ \rho \right]\left[ r \right]\left[ g \right]} = \frac{\left[ {MT}^{- 2} \right]}{\left[ {ML}^{- 3} T^0 \right] \left[ L \right] \left[ {LT}^{- 2} \right]} = \left[ M^0 L^1 T^0 \right] = \left[ L \right]\]

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