Question

A liquid drop placed on a horizontal plane has a near spherical shape (slightly flattened due to gravity). Let $R$ be the radius of its largest horizontal section. A small disturbance causes the drop to vibrate with frequency $v$ about its equilibrium shape. By dimensional analysis the ratio $\frac{v}{\sqrt{\frac{\sigma }{\rho R^3}}}$ can be (Here $\sigma$ is surface tension, $\rho$ is density, $g$ is acceleration due to gravity, and $k$ is an arbitrary dimensionless constant).

• A. $\frac{k\rho g{R}^2}{\sigma }$
• B. $\frac{k\rho R^2}{g\sigma }$
• C. $\frac{k\rho R^3}{g\sigma }$
• D. $\frac{k\rho }{g\sigma }$

Answer 1

Answer: (A)

$\frac{k\rho g{R}^2}{\sigma }$

Surface Tension $\sigma =\frac{Force}{Length}=\frac{mass\times acceleration}{length}=\frac{density\times volume\times acceleration}{length}$

Frequency $v=[T{]}^{-1}$

Dimensionally, Acceleration $\sim \left[g\right]$

Density $\sim \left[\rho \right]$

Length $\sim \left[L\right]=\left[R\right]$

Volume $\sim [L{]}^3=[R^3]$

$\frac{g}{R}\sim \frac{\left[L\right][T{]}^{-2}}{\left[L\right]}=[T{]}^{-2}$

$\sigma \sim \frac{\rho \times R^3\times g}{R}=\left[\rho g{R}^2\right]$ ------ (1)

Therefore,

$\frac{v}{\sqrt{\frac{\sigma }{\rho R^3}}}\sim \frac{[T{]}^{-1}}{[\frac{\rho g{R}^2}{\rho R^3}{]}^{\frac{1}{2}}}=\frac{1}{\left[T\right]\times \sqrt{\frac{g}{R}}}=1$

Thus, the given fraction is $\text{Dimensionless}$

Similarly,

$\frac{\rho g{R}^2}{\sigma }$ is $\text{Dimensionless}$ (from (1))

Thus, by Dimensional Analysis,

$\frac{v}{\sqrt{\frac{\sigma }{\rho R^3}}}\sim \frac{k\rho g{R}^2}{\sigma }$