The time period T of a small drop of liquid (due to surface tension) depends on density ρ, radius r and surface tension S. The relation is-

The correct option is T ∝ (ρr³/S) ⁰.⁵.

To find the correct relationship between the time period $T$ of a drop of liquid and its density $\rho$, radius $r$, and surface tension $S$, we can use the method of dimensional analysis.

First, let us write the dimensional formulas for each of the physical quantities involved:
1. Time period, $\left[T\right]=\left[M^0{L}^0{T}^1\right]$
2. Density, $\left[\rho \right]=\left[M^1{L}^{-3}{T}^0\right]$
3. Radius, $\left[r\right]=\left[M^0{L}^1{T}^0\right]$
4. Surface tension (Force per unit length), $\left[S\right]=\frac{\left[\text{Force}\right]}{\left[\text{Length}\right]}=\frac{\left[ML{T}^{-2}\right]}{\left[L\right]}=\left[M^1{L}^0{T}^{-2}\right]$

Let us assume that the dependency of the time period on these quantities is given by the proportional relationship:
$T\propto {\rho }^a{r}^b{S}^c$
where $a$, $b$, and $c$ are unknown exponents.

Writing this equation in terms of dimensions:
$\left[M^0{L}^0{T}^1\right]={\left[M^1{L}^{-3}{T}^0\right]}^a\times {\left[M^0{L}^1{T}^0\right]}^b\times {\left[M^1{L}^0{T}^{-2}\right]}^c$

Combining the exponents on the right-hand side yields:
$\left[M^0{L}^0{T}^1\right]=\left[M^{a+c}{L}^{-3a+b}{T}^{-2c}\right]$

Comparing the exponents of mass ($M$), length ($L$), and time ($T$) on both sides:
For $T$: $-2c=1\Rightarrow c=-0.5$
For $M$: $a+c=0\Rightarrow a=-c=0.5$
For $L$: $-3a+b=0\Rightarrow b=3a=3\left(0.5\right)=1.5$

Substituting these values back into our proportionality expression:
$T\propto {\rho }^{0.5}{r}^{1.5}{S}^{-0.5}$

This can be simplified by writing the fractional exponents as a single group under a power of 0.5:
$T\propto {\left(\frac{\rho r^3}{S}\right)}^{0.5}$