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 $\frac{v}{\sqrt{\sigma /\rho R^3}}$, can be (Here, a is surface tension,$\rho$ is density, $g$ is acceleration due to gravity and $k$ is an arbitrary dimensionless constant)

• A. $kpg{R}^2/\sigma$
• B. $kp{R}^3/g\sigma$
• C. $kp{R}^2/g\sigma$
• D. $kp/g\sigma$

Answer 1

Answer: (A)

Dimensions of
$\frac{v}{\sqrt{\frac{\sigma }{\rho R^3}}}=\frac{\left[v\right]}{\sqrt{\frac{\left[\sigma \right]}{\left[\rho R^3\right]}}}=\frac{\left[T^{-1}\right]}{\sqrt{\frac{\left[M{T}^{-2}\right]}{\left[\frac{M}{L^3}.L^3\right]}}}$
$=\left[M^0{L}^0{T}^0\right]$
Now, from option $\left(a\right),$
Dimensions of $\frac{k\rho g{R}^2}{\sigma }=\frac{\left[k\rho g{R}^2\right]}{\left[\sigma \right]}$
$\frac{\left[\frac{M}{L^3}\right].\left[L{T}^{-2}\right].\left[L^2\right]}{\left[M{L}^{-2}\right]}$
$=\left[M^0{L}^0{T}^0\right]$
So, by dimensional analysis,