Question

Dimensional methods provide three major advantages in verification, derivation and changing the system of units. Any empirical formula that is derived based on this method has to be verified and proportionality constants found by experimental means. The presence or absence of certain factors - non dimensional constants or variables - cannot be identified by this method. So every dimensionally correct relation cannot be taken as perfectly correct.
The time period of oscillation of a drop depends on surface tenson $\sigma$, density of the liquid $\rho$, and radius r. The relation is

• A. $\sqrt{\frac{rho{r}^2}{\sigma }}$
• B. $\sqrt{\frac{r^2}{\rho \sigma }}$
• C. $\sqrt{\frac{r^3\rho }{\sigma }}$
• D. $\sqrt{\frac{\rho \sigma }{r^3}}$

Answer 1

The correct option is C $\sqrt{\frac{r^3\rho }{\sigma }}$

On checking options

${\left[\frac{r^{3}P}{\sigma }\right]}^{1/2}={\left[\frac{L^3{ML}^{-3}}{{MLT}^{-2}/L}\right]}^{1/3}={\left[T^2\right]}^{1/2}=\left[T\right]$