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Question
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.
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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.
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