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. $kpgR^{2}/\sigma$
• B. $kpR^{3} / g\sigma$
• C. $kpR^{2} / g\sigma$
• D. $kp / g\sigma$

Answer 1

Answer: (A)

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