Question

Test which of the following equations are dimensionally correct. ($S$ = surface tension, 𝜌 = density, $P$ = pressure, $V$ = volume, $\eta$ = coefficient of viscosity, 𝑟 = radius, $v$ = velocity, 𝑡 = time, 𝑙 = length, $H$ = height)

• A. $H=\frac{2S\cos \theta }{\rho rg}$
• B. $v=\sqrt{\frac{P}{\rho }}$
• C. $V=\frac{\pi \rho r^{4}t}{8\eta l}$
• D. All correct

Answer 1

Answer: (A), (B)

$v=\sqrt{\frac{P}{\rho }}$

1) $H=\frac{2S\cos \theta }{\rho rg}$
Here, surface tension is given as force per unit length
$\left[S\right]=\left[F\right]\left[L^{-1}\right]=\left[ML{T}^{-2}\right]=\left[M{T}^{-2}\right]$

$\frac{2S\cos \theta }{\rho rg}=\left[\frac{M{T}^{-2}}{M{L}^{-3}.L.L{T}^{-2}}\right]=\left[L\right]=H$
Hence this expression is dimensionally correct.

2) $v=\sqrt{\frac{P}{\rho }}$
$\left[v\right]=\left[L{T}^{-1}\right]$
$\left[\sqrt{\frac{P}{\rho }}\right]=\frac{[M{L}^{-1}{T}^{-2}{]}^{1/2}}{[M{L}^{-3}{]}^{1/2}}=\left[L{T}^{-1}\right]=\left[v\right]$
Hence this expression is dimensionally correct.

3) $V=\frac{\pi \rho r^{4}t}{8\eta l}$
$\left[V\right]=\left[L^3\right]$
$\left[\eta \right]=\left[M{L}^{-1}{T}^{-1}\right]$
$\therefore \frac{\pi \rho r^{4}t}{8\eta l}=\frac{\left[M{L}^{-3}\right].\left[L^4\right].\left[T\right]}{\left[M{L}^{-1}{T}^{-1}\right].\left[L\right]}=\left[L{T}^2\right]\ne \left[V\right]$
Hence this expression is dimensionally wrong.